Trigonometric Identities

Consider,

(1+cot²A)sin²A

As we know

1+cot²A = cosec²A

Putting the values we get,

(cosec²A)sin²A

As we know,

cosec A = 1/sinA

So,



hence proved

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

so,

(secθ + cosθ ) (secθ - cosθ) = sec²θ - cos²θ ...... (1)

We know,

sec²θ = tan²θ + 1

sin²θ + cos²θ = 1

Use the identities in the eq. (1)

(secθ + cosθ ) (secθ - cosθ) = sec²θ - cos²θ

= (tan²θ + 1) - (1 - sin²θ)

= tan²θ + 1 - 1 + sin²θ

= tan²θ + sin²θ

Hence proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

= RHS
Hence Proved.

tan⁶θ + 3tan²θsec²θ + 1

= (sec²θ - 1)³ + 3(sec²θ - 1)sec²θ + 1 [As, tan²θ = sec²θ - 1]

= (sec⁶θ - 1 - 3sec⁴θ + 3sec²θ) + (3sec⁴θ - 3sec²θ) + 1 [(a + b)³ = a³ - b³ - 3a²b + 3ab²]

= sec⁶θ
= LHS
Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Given:

To prove:

Hence proved

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

To Prove:

Dividing the numerator and denominator by cosθ, we get,




Now, we know that,
sec²θ - tan²θ = 1
Therefore, replacing 1 by sec²θ - tan²θ in the numerator only, we get,

As we know,
a² - b² = (a-b)(a+b)



= secθ + tanθ
Now, multiplying and dividing by secθ - tanθ, we get,

As we know,
sec²θ - tan²θ = 1

= R.H.S
Hence, proved.

Hence Proved.

To prove:

Proof:

Consider LHS,





Use the formula:

secθ = 1/cos θ and tanθ=sinθ/cosθ




Do rationalization,








Use the formula cos²θ + sin²θ = 1








= - tan A

Consider RHS,



Use the formula:

secθ = 1/cos θ and tanθ=sinθ/cosθ




Do rationalization,









Use the formula cos²θ + sin²θ = 1







= - tan A

LHS = RHS

Hence proved.

Hence Proved.

To prove:

Use the formula

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Therefore, LHS = RHS

Hence Proved.

To prove:

Proof:

Consider LHS,

(secA-cosecA)(1+tanA+cotA)

We know,

cosecA=1/sinA,

secA=1/cosA,

tanA=sinA/cosA,

cotA=cosA/sinA

So,





Using the formula a³ - b³ = (a-b) (a²+b²+ab) we get,

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

From (1) and (2) we get

LHS = RHS

Hence Proved.

To Prove:

Proof:

Consider the LHS,




= sinA - cosA+ cotA sinA - cotA cosA + tanA sinA - tanA cosA


Use the formula:













We know:




So,





Again use the formula:






So,









Use the formula:







Therefore,

Hence Proved.

sin²A cos²B - cos²A sin²B= sin²A(1 – sin²B) – (1 – sin²A)sin²B

= sin²A – sin²A sin²B – sin²B + sin²A sin²B

= sin²A – sin²B

= RHS

Hence Proved

Use the formula

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

From (1) and (2), we get

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

(i)

Hence Proved.

(ii)

Hence Proved.

(iii)

Hence Proved.

(iv)

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

Hence Proved.

_Now,

_Now,

_Now,

Given: sec θ + tan θ = x ……………(i)

To find: sec θ

We know that 1 + tan² θ = sec² θ

⇒ sec² θ – tan² θ = 1

∵ a² – b² = (a – b) (a + b)

∴ sec² θ – tan² θ = (sec θ – tan θ) (sec θ + tan θ) = 1

⇒ From (i), we have

⇒ (sec θ – tan θ) x = 1

…………………(ii)

Adding (i) and (ii), we get

An equation that is true for all values of the variables involved is said to be an identity. For example:

a² – b² = (a – b) (a + b)

sin² θ + cos² θ = 1

Given: sec θ + tan θ = x ……………(i)

To find: tan θ

We know that 1 + tan² θ = sec² θ

⇒ sec² θ – tan² θ = 1

∵ a² – b² = (a – b) (a + b)

∴ sec² θ – tan² θ = (sec θ – tan θ) (sec θ + tan θ) = 1

⇒ From (i), we have

⇒ (sec θ – tan θ) x = 1

…………………(ii)

Subtracting (ii) from (i), we get

To find: (1 – cos² θ) cosec² θ

∵

∴

…………(i)

∵ sin² θ + cos² θ = 1

∴ sin² θ = 1 – cos² θ

⇒ from (i), we have

To find: (1 + cot²θ) sin²θ

∵ 1 + cot² θ = cosec² θ

∴ (1 + cot²θ) sin²θ = cosec² θ sin² θ

Also,

Note: Since all the options involve the trigonometric ratios sec θ and tan θ, so we divide the whole term (numerator as well as denominator) by cos θ.

To find:

Consider

Dividing numerator and denominator by cos θ, we get

Rationalizing the term by multiplying it by ,

Now, as 1 + tan² θ = sec² θ

⇒ sec² θ – tan² θ = 1

To find:

∵ 1 + tan² θ = sec² θ

∴

Also, we know that

⇒

⇒

Also,

∵ sin² θ + cos² θ = 1

∴

Note: Since all the options involve the trigonometric ratios cosec θ and cot θ, so we divide the whole term (numerator as well as denominator) by sin θ.

To find:

Consider

Dividing numerator and denominator by sin θ, we get

Rationalizing the term by multiplying it by ,

Now, as 1 + cot² θ = cosec² θ

⇒ cosec² θ – cot² θ = 1

Given: sec θ + tan θ = x ……………(i)

To find: sec θ – tan θ

We know that 1 + tan² θ = sec² θ

⇒ 1 = sec² θ – tan² θ

Now, ∵ a² – b² = (a – b) (a + b)

⇒ 1 = sec² θ – tan² θ = (sec θ – tan θ) (sec θ + tan θ)

⇒ From (i), we have

1 = (sec θ – tan θ) x

Note: Since all the options involve the trigonometric ratio tan θ, so we use the identity 1 + tan² θ = sec² θ.

To find: sec⁴ A – sec² A

Consider sec⁴ A – sec² A = (sec² A)² – sec² A

Now, as sec² A = 1 + tan² A

⇒ sec⁴ A – sec² A = (sec² A)² – sec² A

= (1 + tan² A)² – (1 + tan² A)

= 1 + tan⁴ A + 2 tan² A – 1 – tan² A

= tan⁴ A + tan² A

To find: cos⁴ A – sin⁴ A

Consider cos⁴ A – sin⁴ A = (cos² A)² – (sin² A)²

∵ a² – b² = (a – b) (a + b)

∴ cos⁴ A – sin⁴ A = (cos² A)² – (sin² A)²

= (cos² A – sin² A) (cos² A + sin² A)

= (cos² A – sin² A) [∵ cos² A + sin² A = 1]

= cos² A – (1 – cos² A) [∵ sin²A = 1 – cos²A]

= cos² A – 1 + cos² A = 2 cos² A – 1

Given: cosec θ – cot θ = α ………………(i)

To find: cosec θ + cot θ

We know that 1 + cot² θ = cosec² θ

⇒ 1 = cosec² θ – cot² θ

Now, ∵ a² – b² = (a – b) (a + b)

⇒ 1 = cosec² θ – cot² θ = (cosec θ – cot θ) (cosec θ + cot θ)

⇒ From (i), we have

1 = α (cosec θ + cot θ)

To find: cosec² (90° − θ) − tan²θ

∵ cosec (90° – θ) = sec θ

∴ cosec² (90° – θ) = sec² θ

⇒ cosec² (90° θ) − tan² θ = sec² θ – tan² θ

Now, ∵ 1 + tan² θ = sec² θ

∴ cosec² (90° − θ) − tan² θ = sec² θ – tan² θ

= 1 + tan² θ – tan² θ = 1

To find:

Consider

Rationalizing the above fraction by (1 – cos θ),

[∵ (a – b) (a + b) = a² – b²]

∵ sin² θ + cos² θ = 1

⇒ sin² θ = 1 – cos² θ

To find: sin A cos (90° − A) + cos A sin (90° − A)

∵ cos (90° – A) = sin A and sin (90° – A) = cos A ……………(i)

∴ sin A cos (90° − A) + cos A sin (90° − A)

= sin A sin A + cos A cos A [Using (i)]

= sin² A + cos² A

Now, ∵ sin² θ + cos² θ = 1

∴ sin A cos (90° − A) + cos A sin (90° − A)

= sin² A + cos² A = 1

Given:
To find:The value of
Solution:

Use:

So,

Using the identity,

a² – b² = (a – b) (a + b)

To find:

∵

Also, we know that 1 + cot² θ = cosec² θ

⇒ cot² θ – cosec² θ = – 1

To find: (1 + cot θ − cosec θ) (1 + tan θ + sec θ)

Consider (1 + cot θ − cosec θ) (1 + tan θ + sec θ)

[∵ (a – b) (a + b) = a² – b²]

[∵ sin² θ + cos² θ = 1]

To find:

Consider

∴

[∵ sin² θ = 1 – cos² θ]

Given: x = a sin θ and y = b cos θ

⇒ x² = a² sin² θ and y² = b² cos² θ ………(i)

To find: b²x² + a²y²

Consider b²x² + a²y² = b² a² sin² θ + a² b² cos² θ

= a² b² (sin² θ + cos² θ)

= a² b² (1) [∵ sin² θ + cos² θ = 1]

= a² b²

To find: (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ)

∵

∴ (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ)

Now, as sin² θ + cos ² θ = 1

⇒ sin² θ = 1 – cos² θ

And cos² θ = 1 – sin² θ

Hence the answer is 'B'

Given:

To find: cot θ + cosec θ

∵ sin² θ + cos² θ = 1

∴ cos² θ = 1 – sin² θ

Now, as

Also,

To find: 9 cot² θ – 9 cosec² θ

Consider 9 cot² θ – 9 cosec² θ = 9 (cot² θ – cosec² θ)

Now ∵ 1 + cot² θ = cosec² θ

⇒ cot² θ – cosec² θ = – 1

⇒ 9 cot² θ – 9 cosec² θ = 9 (cot² θ – cosec² θ) = 9 ( – 1) = – 9

Given: x = a sin θ and y = b cos θ

⇒ x² = a² sin² θ and y² = b² cos² θ …………………(i)

To find: b²x² + a²y²

Consider b²x² + a²y² = b² a² sin² θ + a² b² cos² θ

= a² b² (sin² θ + cos² θ)

= a² b² (1) [∵ sin² θ + cos² θ = 1]

= a² b²

To find:

∵

Now, as 1 + tan² θ = sec² θ

⇒ tan² θ – sec² θ = – 1

Given: x = a sec θ and y = b tan θ

⇒ x² = a² sec² θ and y² = b² tan² θ ……………………(i)

To find: b²x² – a²y²

Consider b²x² – a²y² = b² a² sec² θ – a² b² tan² θ

= a² b² (sec² θ – tan² θ)

= a² b² (1) [∵ sec² θ – tan² θ = 1]

= a² b²

To find:

Consider

To find:

We know that 1 + tan² θ = sec² θ

And 1 + cot² θ = cosec² θ

⇒ tan² θ – sec² θ = – 1

And cot² θ – cosec² θ = – 1

To find: 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ)

First, we consider

sin⁶ θ + cos⁶ θ = (sin² θ)³ + (cos² θ)³

Now, as (a + b)³ = a³ + b³ + 3a²b + 3ab²

⇒ a³ + b³ = (a + b)³ – 3a²b – 3ab²

⇒ sin⁶ θ + cos⁶ θ

= (sin² θ)³ + (cos² θ)³

= (sin² θ + cos² θ)³ – 3 (sin² θ)² cos² θ – 3 sin² θ (cos² θ)²

= 1 – 3 sin⁴ θ cos² θ – 3 sin² θ cos⁴ θ [∵ sin² θ + cos² θ = 1]

= 1 – 3 sin² θ cos² θ (sin² θ + cos² θ)

= 1 – 3 sin² θ cos² θ [∵ sin² θ + cos² θ = 1] ………(i)

Next, we consider

sin⁴ θ + cos⁴ θ = (sin² θ)² + (cos² θ)²

Now, as (a + b)² = a² + b² + 2ab

⇒ a² + b² = (a + b)² – 2ab

⇒sin⁴ θ + cos⁴ θ

= (sin² θ)² + (cos² θ)²

= (sin² θ + cos² θ)² – 2 sin² θ cos² θ

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

Now, using (i) and (ii), we have

2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ)

= 2(1 – 3 sin² θ cos² θ) – 3(1 – 2 sin² θ cos² θ)

= 2 – 6 sin² θ cos² θ – 3 + 6 sin² θ cos² θ

= 2 – 3 = – 1

To find: (1 + tan²θ) (1 − sin θ) (1 + sin θ)

∵ (a – b) (a + b) = a² – b²

∴ (1 + tan²θ) (1 − sin θ) (1 + sin θ)

= (1 + tan² θ) (1 – sin² θ)

Now, as sin² θ + cos² θ = 1

⇒ 1 – sin² θ = cos² θ ……………………(i)

Also, we know that 1 + tan² θ = sec² θ …………………(ii)

Using (i) and (ii), we have

(1 + tan²θ) (1 − sin θ) (1 + sin θ)

= (1 + tan² θ) (1 – sin² θ)

= sec² θ cos² θ

∵

⇒ (1 + tan² θ) (1 − sin θ) (1 + sin θ)

= sec² θ cos² θ

=

Given: a cos θ + b sin θ = 4

Squaring both sides, we get

(a cos θ + b sin θ)² = 4²

⇒ a² cos² θ + b² sin² θ + 2ab sin θ cos θ = 16 ………(i)

and a sin θ – b cos θ = 3

Squaring both sides, we get

(a sin θ – b cos θ)² = 3²

⇒ a² sin² θ + b² cos² θ – 2ab sin θ cos θ = 9 ………(ii)

To find: a² + b²

Adding (i) and (ii), we get

a² cos² θ + b² sin² θ + 2ab sin θ cos θ

+ a² sin² θ + b² cos² θ – 2ab sin θ cos θ = 16 + 9

⇒ a² (sin² θ + cos² θ) + b² (sin² θ + cos² θ) = 25

⇒ a² + b² = 25 [∵ sin² θ + cos² θ = 1]

Given:

To find: tan A + cot A

∵ sin² A + cos² A = 1

⇒ sin² A = 1 – cos² A

Now, as

And

Given: a cot θ + b cosec θ = p

Squaring both sides, we get

(a cot θ + b cosec θ)² = p²

⇒ a² cot² θ + b² cosec² θ + 2ab cot θ cosec θ = p² ……(i)

and b cot θ + a cosec θ = q

Squaring both sides, we get

(b cot θ + a cosec θ)² = q²

⇒ b² cot² θ + a² cosec² θ + 2ab cot θ cosec θ = q² ……(ii)

To find: p² – q²

Subtracting (ii) from (i), we get

a² cot² θ + b² cosec² θ + 2ab cot θ cosec θ – b² cot² θ – a² cosec² θ – 2ab cot θ cosec θ = p² – q²

⇒ P² – q² = a² (cot² θ – cosec² θ) + b² (cosec² θ – cot² θ)

= a² ( – 1) + b² (1) [∵1 = cosec² θ – cot² θ]

= b² – a²

Given:

To find: The value of 2 cot² θ + 2.

∵

⇒ cosec² θ = 3² = 9

Also, 1 + cot² θ = cosec² θ

⇒ cot² θ = cosec² θ – 1 = 9 – 1 = 8

⇒ 2 cot² θ + 2 = 2 (8) + 2 = 16 + 2 = 18
Hence, the value of 2 cot² θ + 2 is 18.

To find: sin² 29° + sin²61°

Consider sin² 29° + sin²61°

∵ 29 = 90 – 61

∴ sin² 29° + sin²61° = sin² (90° – 61°) + sin² 61°

Now, as sin (90° – θ) = cos θ

⇒ sin² 29° + sin²61° = sin² (90° – 61°) + sin² 61°

= cos² 61° + sin² 61°

= 1 [sin² θ + cos² θ = 1]

Given:

To find: 9 tan² θ + 9

∵

∴

Also, we know that 1 + tan² θ = sec² θ

Given: x = r sin θ cos ϕ, y = r sin θ sin φ and z = r cos θ,

Solution:

x = r sin θ cos ϕ

Squaring both sides, we get

x² = r² sin² θ cos²ϕ ……….(i)

and y = r sin θ sin ϕ

⇒ y² = r² sin² θ sin²ϕ ……….(ii)

z = r cos θ

Squaring both sides, we get

⇒ z² = r² cos² θ ……….(iii)

Adding (i), (ii) and (iii), we get

x² + y² + z² = r² sin² θ cos²ϕ + r² sin² θ sin²ϕ + r² cos² θ

= r² (sin² θ cos²ϕ + sin² θ sin²ϕ + cos² θ)

= r² [sin² θ (cos²ϕ + sin²ϕ) + cos² θ]

= r² [sin² θ + cos² θ]

= r²

Hence x² + y² + z² = r²

Given: sec²θ (1 + sin θ) (1 − sin θ) = k

To find: k

Consider sec²θ (1 + sin θ) (1 − sin θ)

∵ (a – b) (a + b) = a² – b²

∴ sec²θ (1 + sin θ) (1 − sin θ) = sec² θ (1 – sin² θ)

Now, as sin² θ + cos² θ = 1

⇒ cos² θ = 1 – sin² θ

⇒ sec² θ (1 + sin θ) (1 − sin θ) = sec² θ (1 – sin² θ)

= sec² θ cos² θ

Now, ∵

⇒ sec² θ (1 + sin θ) (1 − sin θ) = sec² θ (1 – sin² θ)

= sec² θ cos² θ

⇒ k = 1

Given: cosec²θ (1 + cos θ) (1 − cos θ) = λ

To find: λ

Consider cosec²θ (1 + cos θ) (1 − cos θ)

∵ (a – b) (a + b) = a² – b²

∴ cosec²θ (1 + cos θ) (1 − cos θ) = cosec² θ (1 – cos² θ)

Now, as sin² θ + cos² θ = 1

⇒ sin² θ = 1 – cos² θ

⇒ cosec²θ (1 + cos θ) (1 − cos θ) = cosec² θ (1 – cos² θ)

= cosec² θ sin² θ

Now, ∵

⇒ cosec² θ (1 + cos θ) (1 − cos θ) = cosec² θ (1 – cos² θ)

= cosec² θ sin² θ

Given: sin θ + sin² θ = 1

⇒ sin θ = 1 – sin² θ = cos² θ [∵ sin² θ + cos² θ = 1]……(i)

⇒ Sin² θ = (cos² θ)² = cos⁴ θ ……(ii)

To find: cos² θ + cos⁴ θ

Consider cos² θ + cos⁴ θ = sin θ + sin² θ [Using (i) and (ii)]

= 1 [Given]

Given: a cos θ + b sin θ = m

Squaring both sides, we get

(a cos θ + b sin θ)² = m²

⇒ a² cos² θ + b² sin² θ + 2ab cos θ sin θ = m² ……(i)

And a sin θ – b cos θ = n

Squaring both sides, we get

(a sin θ – b cos θ)² = n²

⇒ a² sin² θ + b² cos² θ – 2ab sin θ cos θ = n² ……(ii)

To find: a² + b²

Adding (i) and (ii), we get

a² cos² θ + b² sin² θ + 2ab cos θ sin θ + a² sin² θ + b² cos² θ – 2ab sin θ cos θ = m² + n²

⇒ a² (cos² θ + sin² θ) + b² (sin² θ + cos² θ) = m² + n²

⇒ a² + b² = m² + n² [∵ sin² θ + cos² θ = 1]

Given: sin²θ cos²θ (1 + tan²θ) (1 + cot²θ) = λ

To find: λ

We know that 1 + tan² θ = sec² θ

And 1 + cot² θ = cosec² θ

⇒ sin²θ cos²θ (1 + tan²θ) (1 + cot²θ)

= sin² θ cos² θ sec² θ cosec² θ

Now, ∵

And ∵

⇒ sin² θ cos² θ (1 + tan² θ) (1 + cot² θ)

= sin² θ cos² θ sec² θ cosec² θ

⇒ λ = 1

Given: 5x = sec θ

………(i)

And

………(ii)

To find:

Consider [Using (i)]

[Using (ii)]

Now, as 1 + tan² θ = sec² θ

⇒ 1 = sec² θ – tan² θ

Given: cos A + cos² A = 1

⇒ cos A = 1 – cos² A = sin² A [∵ sin² A + cos² A = 1]……(i)

Squaring both sides, we get

⇒ cos² A = (sin² A)² = sin⁴ A ……(ii)

To find: sin² A + sin⁴ A

Consider sin² A + sin⁴ A = cos A + cos² A [From (i) and (ii)]

= 1

Given: x a sec θ cos ϕ

Squaring both sides, we get

x² = a² sec² θ cos²ϕ

and y = b sec θ sin ϕ

Squaring both sides, we get

y² = b² sec² θ sin²ϕ

And z = c tan θ

⇒ z² = c² tan² θ

………(i)

To find:

Consider

= sec² θ cos²ϕ + sec² θ sin²ϕ

= sec² θ (cos²ϕ + sin²ϕ)

= sec² θ [∵ sin²ϕ + cos²ϕ = 1]

= 1 + tan² θ [∵ 1 + tan² θ = sec² θ]

Given: cosec θ = 2x

…………………(i)

And

…………………(ii)

To find:

Consider [Using (i)]

[Using (ii)]

Now, as 1 + cot² θ = cosec² θ

⇒ 1 = cosec² θ – cot² θ

Given: a cos θ – b sin θ = c

To find: a sin θ + b cos θ

Consider a cos θ – b sin θ = c

Squaring both sides, we get

(a cos θ – b sin θ)² = c²

∵ (a – b)² = a² + b² – 2ab

∴ a cos θ – b sin θ = c

⇒ a² cos² θ + b² sin² θ – 2ab sinθ cos θ = c² ……(i)

Now, ∵ sin² θ + cos² θ = 1

∴ sin² θ = 1 – cos² θ and cos² θ = 1 – sin² θ

⇒ From (i), we have

⇒ a² (1 – sin² θ) + b² (1 – cos² θ) – 2ab sin θ cos θ = c²

⇒ a² – a² sin² θ + b² – b² cos² θ – 2ab sin θ cos θ = c²

⇒ a² + b² – (a² sin² θ + b² cos² θ + 2ab sin θ cos θ) = c²

⇒ – (a² sin² θ + b² cos² θ + 2ab sin θ cos θ) = c² – a² – b²

⇒ a² sin² θ + b² cos² θ + 2ab sin θ cos θ = a² + b² – c²

⇒ (a sin θ)² + (b cos θ)² + 2 (a sin θ) (b cos θ) = a² + b² – c²

⇒ (a sin θ + b cos θ)² = a² + b² – c²

⇒

To find: 9 sec² A – 9 tan² A

Consider 9 sec² A – 9 tan² A = 9 (sec² A – tan² A)

∵ 1 + tan² A = sec² A

∴ 9 sec² A – 9 tan² A = 9 (sec² A – tan² A)

= 9 (1 + tan² A – tan² A) = 9

To find: (1 + tan θ + sec θ) (1 + cot θ − cosec θ)

Consider (1 + tan θ + sec θ) (1 + cot θ − cosec θ)

[∵ (a – b) (a + b) = a² – b²]

[∵ sin² θ + cos² θ = 1]

To find: (sec A + tan A) (1 − sin A)

Consider (sec A + tan A) (1 − sin A)

We know that

∵ (a + b) (a – b) = a² – b²

∴

Also, sin² A + cos² A = 1 ⇒ 1 – sin² A = cos² A

To find:

Consider

∵ 1 + tan² A = sec² A and 1 + cot² A = cosec² A

∴