Introduction To Trigonometry And Its Applications

Given: cos A = …eq. 1

d we know that _{tan A =}

We have value of cos A, we need to find value of sin A

Also we know that, sin A = √ (1-cos² A) …eq. 2

(∵, sin² θ +cos² θ =1

⇒ sin² A = 1-cos² A

⇒ sin A = √ (1-cos² A)

Thus,

Substituting eq. 1 in eq. 2, we get

Sin A =

Therefore,

Given: cos A = …eq. 1

d we know that _{tan A =}

We have value of cos A, we need to find value of sin A

Also we know that, sin A = √ (1-cos² A) …eq. 2

(∵, sin² θ +cos² θ =1

⇒ sin² A = 1-cos² A

⇒ sin A = √ (1-cos² A)

Thus,

Substituting eq. 1 in eq. 2, we get

Sin A =

Therefore,

Given: … eq. 1

And we know that, …eq. 2

We need to find the value of cos A.

…eq. 3

(∵, sin² θ +cos² θ =1

⇒ cos² A = 1-sin² A

⇒ cos A = √ (1-sin² A)

Substituting eq. 1 in eq. 3, we get

cos A = √(1-1/4)

=

Substituting values of sin A and cos A in eq. 2, we get

Given: … eq. 1

And we know that, …eq. 2

We need to find the value of cos A.

…eq. 3

(∵, sin² θ +cos² θ =1

⇒ cos² A = 1-sin² A

⇒ cos A = √ (1-sin² A)

Substituting eq. 1 in eq. 3, we get

cos A = √(1-1/4)

=

Substituting values of sin A and cos A in eq. 2, we get

We have,cosec(75°+θ) – sec(15°-θ) - tan(55°+θ) + cot(35°-θ)

= cosec[90°-(15°-θ)] – sec(15°-θ) – tan(55°+θ) + cot[90°-(55°+θ)]

= sec(15°-θ) – sec(15°-θ) - tan(55°+θ) + tan(55°+θ)

(∵, cosec (90°- θ) = sec θ and cot(90°-θ) = tan θ)

= 0

We have,cosec(75°+θ) – sec(15°-θ) - tan(55°+θ) + cot(35°-θ)

= cosec[90°-(15°-θ)] – sec(15°-θ) – tan(55°+θ) + cot[90°-(55°+θ)]

= sec(15°-θ) – sec(15°-θ) - tan(55°+θ) + tan(55°+θ)

(∵, cosec (90°- θ) = sec θ and cot(90°-θ) = tan θ)

= 0

Given: sin 𝛳 =

We know that, sin² θ +cos² θ =1

⇒ sin² A = 1-cos² A

⇒ sin A = √ (1-cos² A

So, cos 𝛳 = √(1-a²/b² ) = √((b²-a²)/b² ) = √(b²-a² )/b

Hence, the answer is √(b² - a² )/b

Given: sin 𝛳 =

We know that, sin² θ +cos² θ =1

⇒ sin² A = 1-cos² A

⇒ sin A = √ (1-cos² A

So, cos 𝛳 = √(1-a²/b² ) = √((b²-a²)/b² ) = √(b²-a² )/b

Hence, the answer is √(b² - a² )/b

Given: cos(α+β) = 0

We can write, cos(α+β)= cos 90° (∵ , cos 90° = 0)

By comparing cosine equation on either sides,

We get(α+β)= 90°

⇒ α = 90°-β

Now we need to reduce sin (α -β )

So, sin(α-β) = sin(90°-β-β) (∵, we have got the value of α, which is α = 90°-β)

= sin(90°-2β)

= cos 2β (∵, sin(90°-θ) = cos θ)

Therefore, sin(α-β) = cos 2β

Given: cos(α+β) = 0

We can write, cos(α+β)= cos 90° (∵ , cos 90° = 0)

By comparing cosine equation on either sides,

We get(α+β)= 90°

⇒ α = 90°-β

Now we need to reduce sin (α -β )

So, sin(α-β) = sin(90°-β-β) (∵, we have got the value of α, which is α = 90°-β)

= sin(90°-2β)

= cos 2β (∵, sin(90°-θ) = cos θ)

Therefore, sin(α-β) = cos 2β

tan 1°. tan 2°.tan 3° …… tan 89° = tan1°.tan 2°.tan 3°…tan 43°.tan 44°.tan 45°.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

(∵ tan 45° = 1)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan(90°-44°).tan(90°-43°)…tan(90°-3°). tan(90°-2°).tan(90°-1°)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.cot 44°.cot 43°…cot 3°.cot 2°.cot 1°

(∵ tan(90°-θ)=cot θ)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.

(∵ tan θ =

= 1

Hence, tan 1°.tan 2°.tan 3° …… tan 89° = 1

tan 1°. tan 2°.tan 3° …… tan 89° = tan1°.tan 2°.tan 3°…tan 43°.tan 44°.tan 45°.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

(∵ tan 45° = 1)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan(90°-44°).tan(90°-43°)…tan(90°-3°). tan(90°-2°).tan(90°-1°)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.cot 44°.cot 43°…cot 3°.cot 2°.cot 1°

(∵ tan(90°-θ)=cot θ)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.

(∵ tan θ =

= 1

Hence, tan 1°.tan 2°.tan 3° …… tan 89° = 1

Given: cos 9∝ = sin ∝ and 9∝<90° i.e. 9α is an acute angle

And we know that, sin(90°-θ) = cos θ by property.

So, we can write cosine in terms of sine using this property,

cos 9∝ = sin (90°-∝)

Thus, sin (90°-9∝) = sin∝ (∵cos 9∝ = sin(90°-9∝) & sin(90°-∝) = sin∝ )

⇒ 90°-9∝ =∝

⇒ 10∝ = 90° (By rearranging)

⇒ ∝ = 9°

We have got the value of ∝ i.e. ∝ = 9°

Putting it in tan 5∝, we get

tan 5∝ = tan (5.9) = tan 45° = 1

∴, tan 5∝ = 1

Given: cos 9∝ = sin ∝ and 9∝<90° i.e. 9α is an acute angle

And we know that, sin(90°-θ) = cos θ by property.

So, we can write cosine in terms of sine using this property,

cos 9∝ = sin (90°-∝)

Thus, sin (90°-9∝) = sin∝ (∵cos 9∝ = sin(90°-9∝) & sin(90°-∝) = sin∝ )

⇒ 90°-9∝ =∝

⇒ 10∝ = 90° (By rearranging)

⇒ ∝ = 9°

We have got the value of ∝ i.e. ∝ = 9°

Putting it in tan 5∝, we get

tan 5∝ = tan (5.9) = tan 45° = 1

∴, tan 5∝ = 1

Given: ∠C = 90°

By the property of triangle, the sum of the three angles is equal to

∠A+∠B+∠C = 180°

⇒ ∠A+∠B+90° = 180° (∵ ∠C = 90° )

⇒ ∠A+∠B = 90°

Thus, cos(A+B) = cos 90° = 0

Given: ∠C = 90°

By the property of triangle, the sum of the three angles is equal to

∠A+∠B+∠C = 180°

⇒ ∠A+∠B+90° = 180° (∵ ∠C = 90° )

⇒ ∠A+∠B = 90°

Thus, cos(A+B) = cos 90° = 0

Given: sin A ₊ sin² A = 1

⇒ sin A = 1 – sin² A

(By Rearranging)

⇒ sin A = cos² A

(∵, sin² θ +cos² θ =1

⇒ cos² θ = 1 – sin² θ)

Squaring both sides, we get

⇒ sin² A = cos⁴ A or cos⁴ A = sin² A

Thus, cos² A + cos⁴ A = sin A + sin² A = 1 (∵ sin A₊sin² A = 1)

∴ cos² A + cos⁴ A = 1

Given: sin A ₊ sin² A = 1

⇒ sin A = 1 – sin² A

(By Rearranging)

⇒ sin A = cos² A

(∵, sin² θ +cos² θ =1

⇒ cos² θ = 1 – sin² θ)

Squaring both sides, we get

⇒ sin² A = cos⁴ A or cos⁴ A = sin² A

Thus, cos² A + cos⁴ A = sin A + sin² A = 1 (∵ sin A₊sin² A = 1)

∴ cos² A + cos⁴ A = 1

Given: sin α = 1/2 and cos β = 1/2

sin α = 1/2 = sin 30° and cos β = 1/2 = cos 60°

Comparing each sine and cosine angles respectively, we get

α = 30° and β = 60°

Thus, α +β = 30°+60° = 90°

Given: sin α = 1/2 and cos β = 1/2

sin α = 1/2 = sin 30° and cos β = 1/2 = cos 60°

Comparing each sine and cosine angles respectively, we get

α = 30° and β = 60°

Thus, α +β = 30°+60° = 90°

Given:

⇒

⇒

(∵cos(90°-θ) = sin θ and sin(90°-θ) = cos θ)

⇒

⇒ 1/1 + 1 = 2

(Since, as by identity, sin² θ +cos² θ =1

So, and sin² 63° + cos² 63° = 1 )

∴

Given:

⇒

⇒

(∵cos(90°-θ) = sin θ and sin(90°-θ) = cos θ)

⇒

⇒ 1/1 + 1 = 2

(Since, as by identity, sin² θ +cos² θ =1

So, and sin² 63° + cos² 63° = 1 )

∴

Given: 4 tan θ =3

⇒ …eq. 1

)

We have to evaluate

Dividing numerator and denominator by cos θ and substituting eq. 1,

We get

∴

Given: 4 tan θ =3

⇒ …eq. 1

)

We have to evaluate

Dividing numerator and denominator by cos θ and substituting eq. 1,

We get

∴

Given: sin θ – cos θ = 0

And we know, tan 45° = 1

So, tan θ = 1 = tan 45°

By comparing above equation, we get θ = 45°

Thus,

Given: sin θ – cos θ = 0

And we know, tan 45° = 1

So, tan θ = 1 = tan 45°

By comparing above equation, we get θ = 45°

Thus,

As we know that, sin(90°-θ) = cos θ by sine property.

So, sin(45°+θ)-cos(45°-θ) = sin[90°-(45°-θ)]-cos(45°-θ)

= cos(45°-θ)-cos(45°-θ) (By using identity)

= 0

As we know that, sin(90°-θ) = cos θ by sine property.

So, sin(45°+θ)-cos(45°-θ) = sin[90°-(45°-θ)]-cos(45°-θ)

= cos(45°-θ)-cos(45°-θ) (By using identity)

= 0

Lets assume AB = 2√3 m, i.e. length of shadow on the ground from the pole

and BC = 6 m, i.e. height of the pole as mentioned in the diagram.

Let sun make an angle of θ on the ground, such that ∠CAB=θ

So in ∆ABC, tan θ = BC/AB

(∵ tan θ = perpendicular/base)

⇒ tan θ = tan 60° (∵ tan 60° = √3)

By comparing, we get

θ = 60°

∴, the sun’s elevation is 60°

Lets assume AB = 2√3 m, i.e. length of shadow on the ground from the pole

and BC = 6 m, i.e. height of the pole as mentioned in the diagram.

Let sun make an angle of θ on the ground, such that ∠CAB=θ

So in ∆ABC, tan θ = BC/AB

(∵ tan θ = perpendicular/base)

⇒ tan θ = tan 60° (∵ tan 60° = √3)

By comparing, we get

θ = 60°

∴, the sun’s elevation is 60°

True

⇒

(∵ tan (90° -θ) = cot θ)

True

⇒

(∵ tan (90° -θ) = cot θ)

False

cos² 23° - sin² 67° =(cos 23°+sin 67°)(cos 23°-sin 67°)

(∵(a²-b²) = (a+b)(a-b))

= [cos 23°+sin(90°-23°)] [cos 23°-sin(90°-23°)]

= (cos 23°+cos 23°)(cos 23°-cos 23°) (∵sin(90°-θ) = cos θ)

= (cos 23°+cos 23°).0

= 0, which is neither positive nor negative

False

cos² 23° - sin² 67° =(cos 23°+sin 67°)(cos 23°-sin 67°)

(∵(a²-b²) = (a+b)(a-b))

= [cos 23°+sin(90°-23°)] [cos 23°-sin(90°-23°)]

= (cos 23°+cos 23°)(cos 23°-cos 23°) (∵sin(90°-θ) = cos θ)

= (cos 23°+cos 23°).0

= 0, which is neither positive nor negative

False

We know,

sin θ increases when 0° ≤ θ ≤ 90°

cos θ decreases when 0° ≤ θ ≤ 90°

And (sin 80°-cos 80°) = (increasing value-decreasing value) which is always equal to a positive value.

∴ (sin 80°-cos 80°) > 0, which is positive (not negative).

False

We know,

sin θ increases when 0° ≤ θ ≤ 90°

cos θ decreases when 0° ≤ θ ≤ 90°

And (sin 80°-cos 80°) = (increasing value-decreasing value) which is always equal to a positive value.

∴ (sin 80°-cos 80°) > 0, which is positive (not negative).

True

LHS: √((1 – cos² θ) sec² θ )

= RHS

True

LHS: √((1 – cos² θ) sec² θ )

= RHS

True

Given: cos A+cos² A = 1

(Rearranging, and taking cos² A on the right side of equation)

⇒ cos A = 1- cos² A

(∵ sin² θ+cos² θ = 1 ⇒ sin² θ = 1- cos² θ)

⇒ cos A = sin² A …eq. 1

Squaring both sides,

we get cos² A = sin⁴ A …eq. 2

We have to find sin²A+sin⁴ A=1

So, adding eq. 1 and eq. 2, we get

sin²A + sin⁴ A= cos A + cos² A (As given)

∴ sin²A+ sin⁴ A = 1

True

Given: cos A+cos² A = 1

(Rearranging, and taking cos² A on the right side of equation)

⇒ cos A = 1- cos² A

(∵ sin² θ+cos² θ = 1 ⇒ sin² θ = 1- cos² θ)

⇒ cos A = sin² A …eq. 1

Squaring both sides,

we get cos² A = sin⁴ A …eq. 2

We have to find sin²A+sin⁴ A=1

So, adding eq. 1 and eq. 2, we get

sin²A + sin⁴ A= cos A + cos² A (As given)

∴ sin²A+ sin⁴ A = 1

_False

_{L.H.S= (tan θ+2)(2 tan θ+1)}

_{Multiplying them,}

= 2 tan² θ + tan θ + 4 tan θ + 2

= 2 tan²θ+5 tan θ+2

= 2(sec² θ-1) +5 tan θ+2

(∵ sec² θ - tan² θ = 1 ⇒ tan²θ = sec² θ-1)

= 2 sec² θ-2+5 tan θ+2

= 5 tan θ+ 2 sec² θ ≠R.H.S

∴, L.H.S ≠ R.H.S

_False

_{L.H.S= (tan θ+2)(2 tan θ+1)}

_{Multiplying them,}

= 2 tan² θ + tan θ + 4 tan θ + 2

= 2 tan²θ+5 tan θ+2

= 2(sec² θ-1) +5 tan θ+2

(∵ sec² θ - tan² θ = 1 ⇒ tan²θ = sec² θ-1)

= 2 sec² θ-2+5 tan θ+2

= 5 tan θ+ 2 sec² θ ≠R.H.S

∴, L.H.S ≠ R.H.S

False

Let us understand this by an example - A tower 2√ 3 m high casts a shadow 2 m long on the ground, then the sun’s elevation is 60°

In ACB,

()

⇒ tan θ = √ 3 = tan 60°

_{∴θ = 60°}

Now, if the same height of tower casts a shadow 4m more than that from preceding length, then the Sun’s elevation becomes 30^°

In APB,

⇒

⇒

⇒

_{∴θ = 30°}

Hence, we can conclude that as the length of the shadow of the tower increases, the angle of elevation of the sun decreases. And the above statement is false.

False

Let us understand this by an example - A tower 2√ 3 m high casts a shadow 2 m long on the ground, then the sun’s elevation is 60°

In ACB,

()

⇒ tan θ = √ 3 = tan 60°

_{∴θ = 60°}

Now, if the same height of tower casts a shadow 4m more than that from preceding length, then the Sun’s elevation becomes 30^°

In APB,

⇒

⇒

⇒

_{∴θ = 30°}

Hence, we can conclude that as the length of the shadow of the tower increases, the angle of elevation of the sun decreases. And the above statement is false.

False

Let P be the point where the man is standing and C be the point where the cloud is. MO = 3 m, the platform’s height from the surface of the lake.

The angles, θ₁= the angle of elevation of the cloud and θ₂ = the angle of depression of the cloud.

The height of reflection of cloud is h+3 because height of lake is also added to the platform’s height.

So, the angle of depression is different in lake to the angle of elevation of cloud above the surface of the lake.

In

Or …eq. 1

In

Or

Or

Or …eq. 2

From eq. 1 and eq. 2,

⇒ θ₂ ≠ θ₁

(∵, is an extra factor, that is why tan θ₁ and tan θ₂ cannot be equal and so θ₁ and θ₂ cannot be equal)

∴ This proves that angle of elevation is not equal to the angle of depression of the sun.

False

Let P be the point where the man is standing and C be the point where the cloud is. MO = 3 m, the platform’s height from the surface of the lake.

The angles, θ₁= the angle of elevation of the cloud and θ₂ = the angle of depression of the cloud.

The height of reflection of cloud is h+3 because height of lake is also added to the platform’s height.

So, the angle of depression is different in lake to the angle of elevation of cloud above the surface of the lake.

In

Or …eq. 1

In

Or

Or

Or …eq. 2

From eq. 1 and eq. 2,

⇒ θ₂ ≠ θ₁

(∵, is an extra factor, that is why tan θ₁ and tan θ₂ cannot be equal and so θ₁ and θ₂ cannot be equal)

∴ This proves that angle of elevation is not equal to the angle of depression of the sun.

False

Given: ‘a’ is a positive number and a≠1

⇒ AM > GM

(Arithmetic Mean (AM) of a list of non- negative real numbers is greater than or equal to the Geometric mean (GM) of the same list)

If a and b be such numbers, then

and GM = √ab

By assuming that statement is be true.

Similarly, AM and GM of a and 1/a are (a+1/a)/2 and √(a.1/a) respectively.

By property, (a+1/a)/2 > √(a.1/a)

⇒ 2 sin θ > 2 (By our assumption)

⇒ sin θ > 1

But -1 ≤ sin θ ≤ 1

∴ Our assumption is wrong and that 2 sin θ cannot be equal to

False

Given: ‘a’ is a positive number and a≠1

⇒ AM > GM

(Arithmetic Mean (AM) of a list of non- negative real numbers is greater than or equal to the Geometric mean (GM) of the same list)

If a and b be such numbers, then

and GM = √ab

By assuming that statement is be true.

Similarly, AM and GM of a and 1/a are (a+1/a)/2 and √(a.1/a) respectively.

By property, (a+1/a)/2 > √(a.1/a)

⇒ 2 sin θ > 2 (By our assumption)

⇒ sin θ > 1

But -1 ≤ sin θ ≤ 1

∴ Our assumption is wrong and that 2 sin θ cannot be equal to

False

Given: a ≠ b and ab > 0

(Because Arithmetic Mean (AM) of a list of non- negative real numbers is greater than or equal to the Geometric mean (GM) of the same list)

⇒ AM > GM

If a and b be such numbers, then

and GM = √ab

By assuming that is true statement.

Similarly, AM and GM of a² and b² will be,

and GM = √(a².b²

So,

(By AM and GM property as mentioned earlier in the answer)

(By our assumption)

But this not possible since, -1 ≤ cos θ ≤ 1

Thus, our assumption is wrong and

False

Given: a ≠ b and ab > 0

(Because Arithmetic Mean (AM) of a list of non- negative real numbers is greater than or equal to the Geometric mean (GM) of the same list)

⇒ AM > GM

If a and b be such numbers, then

and GM = √ab

By assuming that is true statement.

Similarly, AM and GM of a² and b² will be,

and GM = √(a².b²

So,

(By AM and GM property as mentioned earlier in the answer)

(By our assumption)

But this not possible since, -1 ≤ cos θ ≤ 1

Thus, our assumption is wrong and

False

Let height of tower is h and BC = x m.

In ∆ABC, tan 30°= AC/BC = h/x

…eq. 1

Now let height be 2h (doubled)

In

⇒ θ = tan^-1(1.15) < 60°

(∵ tan^-1 (1.15) = 49°)

Hence, the angle of elevation is not doubled when height is doubled.

False

Let height of tower is h and BC = x m.

In ∆ABC, tan 30°= AC/BC = h/x

…eq. 1

Now let height be 2h (doubled)

In

⇒ θ = tan^-1(1.15) < 60°

(∵ tan^-1 (1.15) = 49°)

Hence, the angle of elevation is not doubled when height is doubled.

True

Let height of the tower be h and distance of the point from its foot is x.

Let angle of elevation be θ₁

In

…eq. 1

Now, when both(height and distance) are increased by we get

New height = h+10%of h

New distance = x+10% of x

In ∆PQR,

From eq.1 and eq.2, we get θ₁ = θ₂

Hence, it is true.

True

Let height of the tower be h and distance of the point from its foot is x.

Let angle of elevation be θ₁

In

…eq. 1

Now, when both(height and distance) are increased by we get

New height = h+10%of h

New distance = x+10% of x

In ∆PQR,

From eq.1 and eq.2, we get θ₁ = θ₂

Hence, it is true.

_L.H.S:

Taking L.C.M of the denominators,

_R.H.S

_{Hence proved.}

_L.H.S:

Taking L.C.M of the denominators,

_R.H.S

_{Hence proved.}

L.H.S:

Taking LCM of the denominators,

:R.H.S

Hence proved.

L.H.S:

Taking LCM of the denominators,

:R.H.S

Hence proved.

Given:

And we know,

So

⇒ perpendicular = 3k and base = 4k

By Pythagoras theorem, (hypotenuse)² = (perpendicular)² + (base)²

⇒ (hypotenuse)² = (3k)²+ (4k)² = 9k²+16k² = 25k²

⇒ hypotenuse = 5k

[∵ (hypotenuse)² = 25k²⇒ hypotenuse = √(25k²) = 5k)

Thus, sin A and cos A can be found out.

Multiplying sin A and cos A,

Hence, proved.

Given:

And we know,

So

⇒ perpendicular = 3k and base = 4k

By Pythagoras theorem, (hypotenuse)² = (perpendicular)² + (base)²

⇒ (hypotenuse)² = (3k)²+ (4k)² = 9k²+16k² = 25k²

⇒ hypotenuse = 5k

[∵ (hypotenuse)² = 25k²⇒ hypotenuse = √(25k²) = 5k)

Thus, sin A and cos A can be found out.

Multiplying sin A and cos A,

Hence, proved.

L.H.S: (sin α + cos α) (tan α + cot α)
As we know,
and

: R.H.S

Hence, proved.

L.H.S: (sin α + cos α) (tan α + cot α)
We know,

As we know sin² θ + cos²θ = 1

: R.H.S

Hence, proved.

L.H.S: (√3 + 1) (3 – cot 30°)

= (√3 + 1) (3 – √3) [∵cos 30° = √3]

= (√3 + 1) √3 (√3 - 1) [∵(3 – √3) = √3 (√3 - 1)]

= ((√3)²– 1) √3 [∵ (√3+1)(√3-1) = ((√3)² – 1)]

= (3-1) √3

= 2√3

Similarly solving R.H.S: tan³ 60° - 2 sin 60°

[]

= 3√3 - √3

= 2√3

∴ L.H.S = R.H.S

Hence, proved.

L.H.S: (√3 + 1) (3 – cot 30°)

= (√3 + 1) (3 – √3) [∵cos 30° = √3]

= (√3 + 1) √3 (√3 - 1) [∵(3 – √3) = √3 (√3 - 1)]

= ((√3)²– 1) √3 [∵ (√3+1)(√3-1) = ((√3)² – 1)]

= (3-1) √3

= 2√3

Similarly solving R.H.S: tan³ 60° - 2 sin 60°

[]

= 3√3 - √3

= 2√3

∴ L.H.S = R.H.S

Hence, proved.

L.H.S:

L.H.S:

L.H.S: tan θ + tan (90° - θ)

= tan θ + cot θ

[∵tan (90° - θ) = cot θ]

= sec θ cosec θ

= sec θ sec (90° - θ) [∵cosec θ = sec (90° - θ)]

: R.H.S

Hence, proved.

L.H.S: tan θ + tan (90° - θ)

= tan θ + cot θ

[∵tan (90° - θ) = cot θ]

= sec θ cosec θ

= sec θ sec (90° - θ) [∵cosec θ = sec (90° - θ)]

: R.H.S

Hence, proved.

Given: Height of the pole = h

Length of the shadow = √3 h

Let angle of elevation of the sun = θ

In

Hence, angle of elevation is 30°.

Given: Height of the pole = h

Length of the shadow = √3 h

Let angle of elevation of the sun = θ

In

Hence, angle of elevation is 30°.

Given: √3 tan θ = 1

⇒ tan θ = 1/√3

⇒ θ = tan^-1 (1/√3) [Taking tan inverse]

⇒ θ = 30°

To find value of sin² θ – cos² θ, substitute value of θ

We get, sin² 30° - cos²30°

= (1/2)² – (√3/2)²

= 1/4 - 3/4

= -2/4 = -1/2

Given: √3 tan θ = 1

⇒ tan θ = 1/√3

⇒ θ = tan^-1 (1/√3) [Taking tan inverse]

⇒ θ = 30°

To find value of sin² θ – cos² θ, substitute value of θ

We get, sin² 30° - cos²30°

= (1/2)² – (√3/2)²

= 1/4 - 3/4

= -2/4 = -1/2

Given: length of ladder = 15 m

Angle between the ladder and the top of the wall = 60°

We have to find the height of the wall, h.

In

Thus, the height of the wall is 7.5 m.

Given: length of ladder = 15 m

Angle between the ladder and the top of the wall = 60°

We have to find the height of the wall, h.

In

Thus, the height of the wall is 7.5 m.

(1+tan² θ)(1 – sin θ)(1+sin θ)

= sec² θ (1 – sin θ)(1+sin θ) [∵ 1+tan² θ = sec² θ]

= sec² θ (1 – sin² θ) [∵ by property, (a+b)(a-b) = a²+b²]

= sec² θ cos² θ [∵ 1 – sin² θ = cos²θ]

= 1

(1+tan² θ)(1 – sin θ)(1+sin θ)

= sec² θ (1 – sin θ)(1+sin θ) [∵ 1+tan² θ = sec² θ]

= sec² θ (1 – sin² θ) [∵ by property, (a+b)(a-b) = a²+b²]

= sec² θ cos² θ [∵ 1 – sin² θ = cos²θ]

= 1

Given: 2 sin² θ – cos² θ = 2

⇒ 2 sin² θ – (1 – sin² θ) = 2 [∵ sin² θ + cos² θ = 1 ⇒ cos² θ = 1 – sin² θ]

⇒ 2 sin² θ – 1 + sin² θ = 2

⇒ 3 sin² θ = 3

⇒ sin² θ = 1

⇒ sin θ = 1

⇒ θ = sin^-1 (1)

⇒ θ = 90°

Given: 2 sin² θ – cos² θ = 2

⇒ 2 sin² θ – (1 – sin² θ) = 2 [∵ sin² θ + cos² θ = 1 ⇒ cos² θ = 1 – sin² θ]

⇒ 2 sin² θ – 1 + sin² θ = 2

⇒ 3 sin² θ = 3

⇒ sin² θ = 1

⇒ sin θ = 1

⇒ θ = sin^-1 (1)

⇒ θ = 90°

L.H.S:

= [∵ ]

= [∵ ]

=

= [∵ cos² θ + sin² θ = 1]

= [∵ ]

= 1

:R.H.S

L.H.S:

= [∵ ]

= [∵ ]

=

= [∵ cos² θ + sin² θ = 1]

= [∵ ]

= 1

:R.H.S

Let PQ = 1.5 m, is the height of the observer.

QB = 20.5 m, is the distance of the observer from the tower

AB = 22 m, is the height of the tower

To find θ, the angle of elevation we need to find AM first.

AM = AB – MB

⇒ AM = AB – PQ

⇒ AM = 22 m – 1.5 m = 20.5 m [∵, MB = PQ]

We have the values of PM and AM i.e. 20.5 m and 20.5 m respectively.

In ∆APM,

⇒ θ = 45°

Hence, required angle of elevation of the top of the tower from the eye of the observer is 45°

Let PQ = 1.5 m, is the height of the observer.

QB = 20.5 m, is the distance of the observer from the tower

AB = 22 m, is the height of the tower

To find θ, the angle of elevation we need to find AM first.

AM = AB – MB

⇒ AM = AB – PQ

⇒ AM = 22 m – 1.5 m = 20.5 m [∵, MB = PQ]

We have the values of PM and AM i.e. 20.5 m and 20.5 m respectively.

In ∆APM,

⇒ θ = 45°

Hence, required angle of elevation of the top of the tower from the eye of the observer is 45°

L.H.S: tan⁴θ + tan² θ

Taking tan² θ common, we get

tan² θ (tan² θ + 1)

= tan² θ sec² θ [∵ sec² θ – tan² θ = 1 ⇒ tan² θ + 1 = sec² θ]

= (sec² θ – 1) sec² θ [∵tan² θ = sec² θ – 1]

= sec⁴ θ – sec² θ

: R.H.S

L.H.S: tan⁴θ + tan² θ

Taking tan² θ common, we get

tan² θ (tan² θ + 1)

= tan² θ sec² θ [∵ sec² θ – tan² θ = 1 ⇒ tan² θ + 1 = sec² θ]

= (sec² θ – 1) sec² θ [∵tan² θ = sec² θ – 1]

= sec⁴ θ – sec² θ

: R.H.S

Given: cosec θ + cot θ = p

Squaring on both sides,

Applying component and dividend,

Hence, proved.

Given: cosec θ + cot θ = p

Squaring on both sides,

Applying component and dividend,

Hence, proved.

L.H.S : √(sec² θ + cosec² θ)

: R.H.S

Hence, proved.

L.H.S : √(sec² θ + cosec² θ)

: R.H.S

Hence, proved.

Let PR = h meter, be the height of the tower.

The observer is standing at point Q such that, the distance between the observer and tower is QR = (20+x) m, where

QR = QS + SR = 20 + x

∠PQR = 30°

∠ PSR = θ

In ∆PQR,

Rearranging the terms,

We get 20 +x = √3 h

In ∆PSR,

Since, angle of elevation increases by 15 when the observer moves 20 m towards the tower. We have,

θ = 30° + 15° = 45°

So,

⇒ h = x

Substituting x=h in eq. 1, we get

h = √3 h – 20

⇒ √3 h – h = 20

⇒ h (√3 - 1) = 20

Rationalizing the denominator, we have

= 10 (√3 + 1)

Hence, the required height of the tower is 10 (√3 + 1) meter.

Let PR = h meter, be the height of the tower.

The observer is standing at point Q such that, the distance between the observer and tower is QR = (20+x) m, where

QR = QS + SR = 20 + x

∠PQR = 30°

∠ PSR = θ

In ∆PQR,

Rearranging the terms,

We get 20 +x = √3 h

In ∆PSR,

Since, angle of elevation increases by 15 when the observer moves 20 m towards the tower. We have,

θ = 30° + 15° = 45°

So,

⇒ h = x

Substituting x=h in eq. 1, we get

h = √3 h – 20

⇒ √3 h – h = 20

⇒ h (√3 - 1) = 20

Rationalizing the denominator, we have

= 10 (√3 + 1)

Hence, the required height of the tower is 10 (√3 + 1) meter.

Given: 1+sin² θ = 3 sin θ cos θ

Dividing L.H.S and R.H.S equations with sin² θ,

We get

⇒ cot² θ +1+1 = 3 cot θ [∵, cosec² θ – cot² θ = 1 ⇒ cosec² θ = cot² θ +1]

⇒ cot² θ +2 = 3 cot θ

⇒ cot² θ –3 cot θ +2 = 0

Splitting the middle term and then solving the equation,

⇒ cot² θ – cot θ –2 cot θ +2 = 0

⇒ cot θ(cot θ -1)–2(cot θ +1) = 0

⇒ (cot θ - 1)(cot θ - 2) = 0

⇒ cot θ = 1, 2

Hence, proved.

Given: 1+sin² θ = 3 sin θ cos θ

Dividing L.H.S and R.H.S equations with sin² θ,

We get

⇒ cot² θ +1+1 = 3 cot θ [∵, cosec² θ – cot² θ = 1 ⇒ cosec² θ = cot² θ +1]

⇒ cot² θ +2 = 3 cot θ

⇒ cot² θ –3 cot θ +2 = 0

Splitting the middle term and then solving the equation,

⇒ cot² θ – cot θ –2 cot θ +2 = 0

⇒ cot θ(cot θ -1)–2(cot θ +1) = 0

⇒ (cot θ - 1)(cot θ - 2) = 0

⇒ cot θ = 1, 2

Hence, proved.

Given: sin θ +2 cos θ = 1

Squaring on both sides,

Hence proved.

Given: sin θ +2 cos θ = 1

Squaring on both sides,

Hence proved.

Let BC = s; PC = t

Let height of the tower be AB = h.

∠ABC = θ and ∠APC = 90° - θ

(∵ the angle of elevation of the top of the tower from two points P and B are complementary)

In

In

Multiplying eq. 1 and eq. 2, we get

⇒ h² = st

⇒ h = √st

Hence the height of the tower is √st.

Let BC = s; PC = t

Let height of the tower be AB = h.

∠ABC = θ and ∠APC = 90° - θ

(∵ the angle of elevation of the top of the tower from two points P and B are complementary)

In

In

Multiplying eq. 1 and eq. 2, we get

⇒ h² = st

⇒ h = √st

Hence the height of the tower is √st.

Let SQ = h be the tower.

∠SPQ = 30° and ∠SRQ = 60°

According to the question, the length of shadow is 50 m long hen angle of elevation of the sun is 30° than when it was 60°. So,

PR = 50 m and RQ = x m

So in ∆SRQ, we have

In ∆SPQ,

Substituting the value of x in the above equation, we get

⇒ 50√3+h = 3h

⇒ 50√3 = 3h - h

⇒ 3h - h = 50√3

⇒ 2h = 50√3

⇒ h = (50√3)/2

⇒ h = 25√3

Hence, the required height is 25√3 m.

Let SQ = h be the tower.

∠SPQ = 30° and ∠SRQ = 60°

According to the question, the length of shadow is 50 m long hen angle of elevation of the sun is 30° than when it was 60°. So,

PR = 50 m and RQ = x m

So in ∆SRQ, we have

In ∆SPQ,

Substituting the value of x in the above equation, we get

⇒ 50√3+h = 3h

⇒ 50√3 = 3h - h

⇒ 3h - h = 50√3

⇒ 2h = 50√3

⇒ h = (50√3)/2

⇒ h = 25√3

Hence, the required height is 25√3 m.

Given that a vertical flag staff of height h is surmounted on a vertical tower of height H(say),

such that FP = h and FO = H.

The angle of elevation of the bottom and top of the flag staff on the plane is ∠PRO = α and ∠FRO = β respectively.

In ∆PRO, we have

And in ∆FRO, we have

Comparing eq. 1 and eq. 2,

Solving for H,

⇒ H tan β = (h+H) tan α

⇒ H tan β – H tan α = h tan α

⇒ H (tan β – tan α) = h tan α

Hence, proved.

Given that a vertical flag staff of height h is surmounted on a vertical tower of height H(say),

such that FP = h and FO = H.

The angle of elevation of the bottom and top of the flag staff on the plane is ∠PRO = α and ∠FRO = β respectively.

In ∆PRO, we have

And in ∆FRO, we have

Comparing eq. 1 and eq. 2,

Solving for H,

⇒ H tan β = (h+H) tan α

⇒ H tan β – H tan α = h tan α

⇒ H (tan β – tan α) = h tan α

Hence, proved.

Given: tan θ+ sec θ = l …eq. 1

_{Multiplying and dividing by (sec θ – tan θ) on numerator and denominator of L.H.S,}

Adding eq. 1and eq. 2, we get

_{Hence, proved.}

Given: tan θ+ sec θ = l …eq. 1

_{Multiplying and dividing by (sec θ – tan θ) on numerator and denominator of L.H.S,}

Adding eq. 1and eq. 2, we get

_{Hence, proved.}

Given that sin θ + cos θ = p and sec θ + cosec θ = q

Taking sec θ + cosec θ = q

Squaring sin θ + cos θ = p,

We have (sin θ + cos θ)² = p²

⇒ sin² θ + cos² θ + 2 sin θ cos θ = p²

⇒ 1+2 sin θ cos θ = p² [∵,sin² θ + cos² θ = 1]

⇒ q+2p = p² q

⇒ q (p² – 1) = 2p

Hence, proved.

Given that sin θ + cos θ = p and sec θ + cosec θ = q

Taking sec θ + cosec θ = q

Squaring sin θ + cos θ = p,

We have (sin θ + cos θ)² = p²

⇒ sin² θ + cos² θ + 2 sin θ cos θ = p²

⇒ 1+2 sin θ cos θ = p² [∵,sin² θ + cos² θ = 1]

⇒ q+2p = p² q

⇒ q (p² – 1) = 2p

Hence, proved.

Given: a sin θ +b cos θ = c

_{Squaring the above equation,}

⇒ a²sin² θ +b² cos² θ + 2ab sin θ cos θ = c^2

Hence, proved.

Given: a sin θ +b cos θ = c

_{Squaring the above equation,}

⇒ a²sin² θ +b² cos² θ + 2ab sin θ cos θ = c^2

Hence, proved.

L.H.S:

Rationalizing it,

: R.H.S

Hence, proved.

L.H.S:

Rationalizing it,

: R.H.S

Hence, proved.

Given that height of one of the tower is 30 m, ∠QAB = 60° and ∠PBA = 30°

Let height of another tower be h m & distance between the towers be x m.

We need to find x and h.

So, in ∆QAB,

& in ∆PBA,

we have got the value of x, i.e. 10√3 m. So, putting the value of x in the above equation,

⇒ h = 10

Thus, we have required distance between the towers, i.e. 10√3 m

& height of another tower, i.e. 10 m.

Given that height of one of the tower is 30 m, ∠QAB = 60° and ∠PBA = 30°

Let height of another tower be h m & distance between the towers be x m.

We need to find x and h.

So, in ∆QAB,

& in ∆PBA,

we have got the value of x, i.e. 10√3 m. So, putting the value of x in the above equation,

⇒ h = 10

Thus, we have required distance between the towers, i.e. 10√3 m

& height of another tower, i.e. 10 m.

Given: the height of tower is h m.

∠ABD = α & ∠ACD = β

Let CD = y and BC = x

In ∆ABD,

In ∆ACD,

Comparing eq. 1 and eq. 2,

⇒ x = h (cot α – cot β)

Hence, we have got the required distance between the two points, i.e. h (cot α – cot β)

Given: the height of tower is h m.

∠ABD = α & ∠ACD = β

Let CD = y and BC = x

In ∆ABD,

In ∆ACD,

Comparing eq. 1 and eq. 2,

⇒ x = h (cot α – cot β)

Hence, we have got the required distance between the two points, i.e. h (cot α – cot β)

Let QO = x and given that BQ = q, such that BO = q+x is the height of the wall.

Let AO = y and given that SA = q.

∠BAO = α and ∠QSO = β

In ∆BAO,

⇒ BO = AB sin α …eq. 1

⇒ AO = AB cos α …eq. 2

In ∆QSO,

⇒ QO = SQ sin β …eq. 3

⇒ SO = SQ cos β …eq. 4

Subtracting eq. 2 from eq. 4, we get

SO –AO = SQ cos β – AB cos α

⇒ SA = SQ cos β – AB cos α

[from the above figure, SO –AO = SA = p]

⇒ p = AB cos β – AB cos α

[∵ SQ=AB=length of the ladder]

⇒ p = AB (cos β – cos α) …eq. 5

And subtracting eq. 3 from eq. 1, we get

BO –QO = AB sin α – SQ sin β

⇒ BQ = AB sin α – SQ sin β

[from the above figure, BO –QO = BQ = q]

⇒ q = AB sin α – AB sin β

[∵ SQ=AB=length of the ladder]

⇒ q = AB (sin α – sin β) …eq. 6

Dividing eq. 5 and eq. 6, we get

Hence, proved.

Let QO = x and given that BQ = q, such that BO = q+x is the height of the wall.

Let AO = y and given that SA = q.

∠BAO = α and ∠QSO = β

In ∆BAO,

⇒ BO = AB sin α …eq. 1

⇒ AO = AB cos α …eq. 2

In ∆QSO,

⇒ QO = SQ sin β …eq. 3

⇒ SO = SQ cos β …eq. 4

Subtracting eq. 2 from eq. 4, we get

SO –AO = SQ cos β – AB cos α

⇒ SA = SQ cos β – AB cos α

[from the above figure, SO –AO = SA = p]

⇒ p = AB cos β – AB cos α

[∵ SQ=AB=length of the ladder]

⇒ p = AB (cos β – cos α) …eq. 5

And subtracting eq. 3 from eq. 1, we get

BO –QO = AB sin α – SQ sin β

⇒ BQ = AB sin α – SQ sin β

[from the above figure, BO –QO = BQ = q]

⇒ q = AB sin α – AB sin β

[∵ SQ=AB=length of the ladder]

⇒ q = AB (sin α – sin β) …eq. 6

Dividing eq. 5 and eq. 6, we get

Hence, proved.

Given that ∠TQO = 60° and ∠TAB = 45°, when AP = 10 m.

So, OB = 10 m from the diagram.

Let PO = x m & if height of the vertical tower, TO = H m, then TB = TO –OB = (H -10) m

In ∆TPO,

In ∆TAB,

⇒ AB = H -10

[∵ TO –TH = H-10]

⇒ x = H -10 …eq. 2

Comparing eq. 1 and eq. 2, we get

Rationalizing it,

Hence, required height of the tower is 5√3 (√3+1) m.

Given that ∠TQO = 60° and ∠TAB = 45°, when AP = 10 m.

So, OB = 10 m from the diagram.

Let PO = x m & if height of the vertical tower, TO = H m, then TB = TO –OB = (H -10) m

In ∆TPO,

In ∆TAB,

⇒ AB = H -10

[∵ TO –TH = H-10]

⇒ x = H -10 …eq. 2

Comparing eq. 1 and eq. 2, we get

Rationalizing it,

Hence, required height of the tower is 5√3 (√3+1) m.

Let WB = h m, height of the smaller window & QO = H m, height of the other window.

Let BO = x m, distance between the two windows.

∠QWM = α & ∠WOB = β

In ∆WOB,

In ∆QWM,

Comparing eq. 1 and eq. 2, we get

(H – h)/tan α = h/tan β

⇒ (H – h) tan β = h tan α

⇒ H tan β – h tan β = h tan α

⇒ H tan β = h tan α +h tan β

⇒ H tan β = h (tan α + tan β)

⇒ H = (h (tan α + tan β))/tan β

⇒ H = h (tan α/tan β + 1)

⇒ H = h (1+tan α/tan β)

⇒ H = h (1+tan α cot β)

Hence, proved.

Let WB = h m, height of the smaller window & QO = H m, height of the other window.

Let BO = x m, distance between the two windows.

∠QWM = α & ∠WOB = β

In ∆WOB,

In ∆QWM,

Comparing eq. 1 and eq. 2, we get

(H – h)/tan α = h/tan β

⇒ (H – h) tan β = h tan α

⇒ H tan β – h tan β = h tan α

⇒ H tan β = h tan α +h tan β

⇒ H tan β = h (tan α + tan β)

⇒ H = (h (tan α + tan β))/tan β

⇒ H = h (tan α/tan β + 1)

⇒ H = h (1+tan α/tan β)

⇒ H = h (1+tan α cot β)

Hence, proved.

Given that the height of lower window is OR = 2m

Height of upper window is RQ = 4 m

Let height of balloon above the ground is BO = H meter & PO = x meter

∠Bw₂R = 60° and ∠Bw₁Q = 30°

In ∆Bw₂R,

In ∆Bw₁Q,

⇒ x = √3 (H - 6) …eq. 2

Comparing eq. 1 and eq. 2, we get

⇒ H – 2 = 3 (H – 6)

⇒ H – 2 = 3H – 18

⇒ 3H – H = 18 – 2

⇒ 2H = 16

⇒ H = 8

Hence, height of the balloon above the ground is 8 meter.

Given that the height of lower window is OR = 2m

Height of upper window is RQ = 4 m

Let height of balloon above the ground is BO = H meter & PO = x meter

∠Bw₂R = 60° and ∠Bw₁Q = 30°

In ∆Bw₂R,

In ∆Bw₁Q,

⇒ x = √3 (H - 6) …eq. 2

Comparing eq. 1 and eq. 2, we get

⇒ H – 2 = 3 (H – 6)

⇒ H – 2 = 3H – 18

⇒ 3H – H = 18 – 2

⇒ 2H = 16

⇒ H = 8

Hence, height of the balloon above the ground is 8 meter.