Trigonometric Identities

Consider the left – hand side:

L.H.S. = (1 – cos²θ) × cosec²θ

= (sin²θ) × cosec²θ (∵ sin²θ + cos² θ = 1)

= 1

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. = (1 + cot²θ) × sin² θ

= (cosec² θ) × sin² θ (∵ 1 + cot² θ = cosec² θ)

= 1

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. = (sec² θ – 1) × cot² θ

= (tan²θ) × cot²θ (∵ 1 + tan² θ = sec² θ)

= 1

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. = (sec² θ – 1)(cosec² θ – 1)

= (tan²θ) × cot²θ (∵ 1 + tan² θ = sec² θ and 1 + cot² θ = cosec² θ)

= 1

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. = (1 – cos² θ) sec² θ

= (sin²θ) × (1/cos²θ) (∵ sin² θ + cos² θ = 1

= tan² θ

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. =

= (sin²θ) + (1/sec² θ) (∵ 1 + tan² θ = sec² θ)

= (sin²θ) + (cos²θ) (∵ sin² θ + cos² θ = 1)

= 1

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. =

= (1/sec²θ) + (1/cosec² θ) (∵ 1 + tan² θ = sec² θ and 1 + tan² θ = sec² θ)

= (cos² θ) + (sin²θ) (∵ sin² θ + cos² θ = 1)

= 1

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. = (1 + cos θ)(1 – cos θ)(1 + cot² θ)

= (1 – cos² θ) × cosec² θ (∵ 1 + cot² θ = cosec² θ)

= (sin² θ) × cosec² θ (∵ sin²θ + cos² θ = 1)

= 1

= R.H.S.

Hence, proved.

To prove: (cosec θ)(1 + cos θ)(cosec θ – cot θ) = 1

Proof:

Consider the left – hand side:

(cosec θ) (1 + cos θ)(cosec θ – cot θ)


⇒ (cosec θ) (1 + cos θ)(cosec θ – cot θ) = (cosec θ + cosec θ cos θ)(cosec θ – cot θ)


since cosec θ = 1/sin θ

⇒ (cosec θ) (1 + cos θ)(cosec θ – cot θ)=


Also



So,

⇒ (cosec θ) (1 + cos θ)(cosec θ – cot θ) =(cosec θ + cot θ)(cosec θ – cot θ)

Use the formula (a + b)(a - b) = a² - b²

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

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. = cot² θ –

=

=

= ( – sin² θ) × sin² θ (∵ sin²θ + cos² θ = 1)

= – 1

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. = tan² θ –

=

=

= ( – cos² θ) × cos² θ (∵ sin²θ + cos² θ = 1)

= – 1

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. = cos² θ +

= cos² θ + (∵ 1 + cot² θ = cosec² θ)

= cos² θ + sin² θ

= ( – cos² θ) × cos² θ (∵ sin²θ + cos² θ = 1)

= – 1

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. =

=

=

= 2 sec² θ

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. = sec θ(1 – sin θ)(sec θ + tan θ)

= × (1 – sin θ) ×

= × (1 – sin θ) ×

=

=

= 1

= R.H.S.

Hence, proved.

To prove: sin θ (1 + tan θ) + cos θ (1+ cot θ) = (sec θ + cosec θ)
Proof:
Consider the left – hand side:

L.H.S. = sin θ (1 + tan θ) + cos θ (1+ cot θ)

= sin θ (1 + ) + cos θ (1+ )

= sin θ + (cos θ) ×

= (cos θ + sin θ)

= (cos θ + sin θ)
We know cos²θ + sin²θ = 1

=

=

= cosec θ + sec θ

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. =

=

= 1+

=

=

=

= 1/sin θ

= cosec θ

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. =

=

=

=

=

=

= 1/cos θ

= sec θ

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. =

=

=

=

= tan θ

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. =

=

=

=

=

= 1

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. =

=

As sin²θ + cos²θ = 1, we have

=

=

= 2/sin θ

= 2cosec θ

= R.H.S.

Hence, proved.

Consider L.H.S. =

=

= +

=

=

=

= sec θ cosec θ + 1

= R.H.S.

Hence, proved.

Consider the left – hand side:

L.H.S. =

=

=

=

=

= cos² θ + sin² θ + sin θ cos θ

= 1 + cos θ sin θ

= R.H.S.

Hence, proved.

=

=

=

=

= cos θ + sin θ

= R.H.S.

Hence, proved.

Consider L.H.S. = (1 + tan² θ) (1 + cot² θ)

= (sec² θ)(cosec² θ)

=

=

=

= R.H.S.

Hence, proved.

Consider L.H.S. =

=

=

=

=

= sin θ (cos³ θ) + cos θ (sin³ θ)

= sin θ cos θ (cos² θ + sin² θ)

= sin θ cos θ

= R.H.S.

Hence, proved.

Consider L.H.S. = sin⁶ θ + cos⁶ θ

= (sin² θ)³ + (cos² θ)³

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

[Using a³ + b³ = (a + b)(a² + b² – ab)]

= ( sin⁴ θ + cos⁴ θ – sin² θ cos² θ)

(∵sin² θ + cos² θ = 1)

= [{(sin² θ + cos² θ)² – 2 sin² θ cos² θ) – sin² θ cos² θ]

(∵(a² + b² ) = (a + b)² – 2ab) )

= [1 – 2 sin² θ cos² θ – sin² θ cos² θ]

= 1 – 3 sin² θ cos² θ

= R.H.S.

Hence, proved.

Consider L.H.S. = sin² θ + cos⁴ θ

= (sin² θ) + (cos² θ)²

= (sin² θ) + (1 – sin² θ)²

= (sin² θ) + 1 + sin⁴ θ – 2sin² θ

= 1 – sin² θ + sin⁴ θ

= cos² θ + sin⁴ θ

= R.H.S.

Hence, proved.

Consider L.H.S. = cosec⁴ θ – cosec² θ

= (cosec² θ)² – (cosec² θ)

= (1 + cot² θ)² – (cosec² θ)

= 1 + cot⁴ θ + 2cot² θ – (cosec² θ)

= 1 + cot⁴ θ + cot² θ – (cosec² θ – cot² θ)

= 1 + cot⁴ θ + cot² θ – 1

= cot⁴ θ + cot² θ

= R.H.S.

Hence, proved.

Consider L.H.S. =

=

=

=

= cos² θ – sin² θ

= R.H.S.

Hence, proved.

Consider L.H.S. =

=

=

=

= sin² θ/ cos² θ

= tan² θ

= R.H.S.

Hence, proved.

Consider L.H.S. = +

=

=

=

=

= [2 (1/cos θ)]/[sin θ /cos θ]

= [2/sin θ]

= 2 cosec θ

= R.H.S.

Hence, proved.

Consider L.H.S. = +

=

=

=

=

=

= 2 cosec θ/cot θ

= 2 (1/sin θ)/(cos θ/sin θ)

= 2/ cos θ

= 2 sec θ

= R.H.S.

Hence, proved.

Consider L.H.S. =

Multiply and divide by (sec θ + 1):

=

=

=

=

=

=

= R.H.S.

Hence, proved.

Consider L.H.S. =

Multiply and divide by (sec θ + tan θ):

=

=

=

=

=

= R.H.S.

Hence, proved.

Consider L.H.S. =

Multiply and divide by (1 + sin θ):

=

=

=

= (1 + sin θ)/cos θ

= (1/cos θ) + (sin θ/cos θ)

= sec θ + tan θ

= R.H.S.

Hence, proved.

Consider L.H.S. =

Multiply and divide by (1 – cos θ):

=

=

=

= (1 – cos θ)/sin θ

= (1/sin θ) – (cos θ/sin θ)

= cosec θ – cot θ

= R.H.S.

Hence, proved.

Consider L.H.S. =

Multiply and divide by (1 + cos θ) in first part and (1 – cos θ) in the second part:

=

=

=

= [(1 + cos θ)/sin θ] + [(1 – cos θ)/sin θ]

= [(1/sin θ) + (cos θ/sin θ)] + [(1/sin θ) – (cos θ/sin θ)]

= [cosec θ + cot θ] + [cosec θ – cot θ]

= 2 cosec θ

= R.H.S.

Hence, proved.

Consider L.H.S. =

Using identities (a³ + b³) = (a + b)(a² + b² – ab) and (a³ – b³) = (a – b)(a² + b² + ab)

∴ L.H.S. =

= (cos² θ + sin² θ – cos θ sin θ) + (cos² θ + sin² θ + cos θ sin θ)

= (1 – cos θ sin θ) + (1 + cos θ sin θ)

= 2

= R.H.S.

Hence, proved.

Consider L.H.S. =

=

= R.H.S.

Hence, proved.

Consider L.H.S. =

=

=

=

=

= R.H.S.

Hence, proved.

Consider L.H.S. =

=

=

=

=

= R.H.S.

Hence, proved.

Consider L.H.S. =

=

=

=

= cos θ/sin θ

= cot θ

= R.H.S.

Hence, proved.

Consider L.H.S. =

Multiply and divide by (cosec θ + cot θ):

=

=

= (cosec θ + cot θ)²

Thus, proved.

Also, consider (cosec θ + cot θ)² = cosec² θ + cot² θ+ 2 cosec θ cot θ

= 1 + cot² θ + cot² θ+ 2 cosec θ cot θ (∵ 1 + cot² θ = cosec² θ)

= (1 + 2 cot² θ + 2 cosec θ cot θ)

= R.H.S.

Hence, proved.

Consider L.H.S. =

Multiply and divide by (sec θ + tan θ):

=

=

= (sec θ + tan θ)²

Thus, proved.

Also, consider (sec θ + tan θ)² = sec² θ + tan² θ+ 2 sec θ tan θ

= 1 + tan² θ + tan² θ+ 2 sec θ tan θ (∵ 1 + tan² θ = sec² θ)

= (1 + 2 tan² θ + 2 sec θ tan θ)

= R.H.S.

Hence, proved.

Consider L.H.S. =

Multiply and divide by ((1 + cos θ) + sin θ):

=

=

=

=

=

=

=

= R.H.S.

Thus, proved.

Consider L.H.S. =

Multiply and divide by (cos θ + 1) + sin θ):

=

=

=

=

=

= R.H.S.

Thus, proved.

Consider L.H.S. =

=

=

= sin θ cos θ ×

= sin θ cos θ ×

= sin θ cos θ ×

= sin θ cos θ ×

= sin θ cos θ ×

= sin θ cos θ/ sin θ cos θ

= 1

= R.H.S.

Consider L.H.S. =

=

=

=

Thus, prove.

Also, consider =

=

= R.H.S.

Hence, proved.

Consider L.H.S. =

=

=

=

=

= (1/sin θ) – (1/cos θ)

= cosec θ – sec θ

= R.H.S.

Hence, proved.

Consider L.H.S. = (1 + tan θ + cot θ)(sin θ – cos θ)

= sin θ – cos θ + tan θ sin θ – tan θ cos θ + cot θ sin θ – cot θ cos θ

= sin θ – cos θ + tan θ sin θ – sin θ + cos θ – cot θ cos θ

= tan θ sin θ – cot θ cos θ

Hence, proved.

Consider L.H.S. = +

= +

= +

= +

= +

= –

= 0

= R.H.S.

Hence, proved.

Consider L.H.S. = (sin² θ cos² θ)

= (sin² θ cos² θ)

= (sin² θ cos² θ)

= (sin² θ cos² θ)

= (sin² θ cos² θ)

= (sin² θ cos² θ)

=

=

=

=

= R.H.S.

Consider the left – hand side =

=

=

=

=

= 0

= R.H.S.

Hence, proved.

Consider the L.H.S. =

=

=

=

= tan A tan B

= R.H.S.

Hence, proved.

If the given equation is an identity, then it is true for every value of θ.

So, let θ = 60°

So, for θ = 60°, consider the L.H.S. = cos² 60° + cos 60°

= (1/2)² + (1/2)

= (1/4) + (1/2)

= 3/4 ≠ 1

Therefore, L.H.S. ≠ R.H.S.

Thus, the given equation is not an identity.

If the given equation is an identity, then it is true for every value of θ.

So, let θ = 30°

So, for θ = 30°, consider the L.H.S. = sin² 30° + sin 30°

= (1/2)² + (1/2)

= (1/4) + (1/2)

= 3/4 ≠ 2

Therefore, L.H.S. ≠ R.H.S.

Thus, the given equation is not an identity.

If the given equation is an identity, then it is true for every value of θ.

So, let θ = 30°

So, for θ = 30°, consider the L.H.S. = tan² 30° + sin 30°

= (1/√3)² + (1/2)

= (1/3) + (1/2)

= 5/6

Consider the R.H.S. = cos² 30° = (√3/2)²

= 3/4

Therefore, L.H.S. ≠ R.H.S.

Thus, the given equation is not an identity.

Consider R.H.S. = (2cos³ θ – cos θ) tan θ

= cos θ(2cos² θ – 1)

= (2cos² θ – 1)sin θ

Consider L.H.S. = (sin θ – 2 sin³ θ)

= sin θ(1 – 2 sin² θ)

= sin θ[1 – 2(1 – cos² θ)]

= sin θ [1 – 2 + 2cos² θ]

= sin θ (2cos² θ – 1)

Therefore, L.H.S. = R.H.S.

Hence, proved.

Given: a cos θ + b sin θ = m …….(1)

a sin θ – b cos θ = n …….(2)

Square equation (1) and (2) on both sides:

a² cos² θ + b² sin² θ + 2ab cos θ sin θ = m² …….(3)

a² sin² θ + b² cos² θ – 2ab cos θ sin θ = n² ……..(4)

Add equation (3) and (4):

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

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

⇒ a^{2 +} b² = m² + n²

Hence, proved.

Given: a sec θ + b tan θ = x …….(1)

a tan θ + b sec θ = y …….(2)

Square equation (1) and (2) on both sides:

a² sec² θ + b² tan² θ + 2ab sec θ tan θ = x² …….(3)

a² tan² θ + b² sec² θ + 2ab sec θ tan θ = y² ……..(4)

Subtract equation (4) from (3):

[a² sec² θ + b² tan² θ + 2ab sec θ tan θ] – [a² tan² θ + b² sec² θ + 2ab sec θ tan θ] = x² – y²

⇒ a² (sec² θ – tan² θ) + b² (tan² θ – sec² θ) = x² – y²

⇒ a² – b² = x² – y² (∵sec² θ = 1 + tan² θ)

Hence, proved.

Given: sin θ – cos θ = 1 …….(1)

cos θ + sin θ = 1 …….(2)

Square equation (1) and (2) on both sides:

sin² θ + cos² θ – 2 cos θ sin θ = 1 …….(3)

cos² θ + sin² θ + 2 cos θ sin θ = 1 ……..(4)

Add equation (3) and (4):

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

⇒ (1) + (1) = 2

⇒ + = 2

Hence, proved.

Given: (sec θ + tan θ) = m …………….(1)

(sec θ – tan θ) = n …………….(2)

Multiply equation (1) and (2):

(sec θ + tan θ) (sec θ – tan θ) = mn

(sec² θ – tan² θ) = mn

1 = mn (∵ 1 + tan² θ = sec² θ)

Therefore, mn = 1.

Hence, proved.

Given: (cosec θ + cot θ) = m …………….(1)

(cosec θ – cot θ) = n …………….(2)

Multiply equation (1) and (2):

(cosec θ + cot θ) (cosec θ – cot θ) = mn

(cosec² θ – cot² θ) = mn

1 = mn (∵ 1 + cot² θ = cosec² θ)

Therefore, mn = 1.

Hence, proved.

Given: x = a cos³ θ

y = b sin³ θ

Consider L.H.S. =

=

= (cos³ θ)^2/3 + (sin³ θ)^2/3

= (cos² θ + (sin² θ)

= 1 = R.H.S.

Hence, proved.

Given: tan θ + sin θ = m …….(1)

tan θ – sin θ = n …….(2)

Square equation (1) and (2) on both sides:

tan² θ + sin² θ + 2 sin θ tan θ = m² …….(3)

tan² θ + sin² θ – 2 sin θ tan θ = n² ……..(4)

Subtract equation (4) from (3):

[tan² θ + sin² θ + 2 sin θ tan θ] – [tan² θ + sin² θ – 2 sin θ tan θ] = m² – n²

⇒ 4sin θ tan θ = m² – n²

Square both sides:

⇒ 16 sin² θ tan² θ = (m² – n²)²

Therefore, (m² – n²)² = 16 sin² θ tan² θ

Also, 16mn = 16 × (tan θ + sin θ) × (tan θ – sin θ)

= 16 (tan² θ – sin² θ)

= 16[(sin² θ/ cos² θ) – sin² θ]

= 16[sin² θ ]

= 16 sin² θ(sin² θ/ cos² θ)

= 16 sin² θ tan² θ

Therefore, (m² – n²)² = 16mn

Hence, proved.

Given: (cot θ + tan θ) = m

(sec θ – cos θ) = n

Since, m = cot θ + tan θ

= (1/tan θ) + tan θ

=

= sec² θ/tan θ

= 1/(sin θ cos θ)

Also, n = sec θ – cos θ

= (1/cos θ) – cos θ

= (1 – cos² θ)/cos θ

= sin² θ/cos θ

Now, consider the left – hand side:

(m²n) ^2/3 – (mn²) ^2/3 =

=

=

=

= (1 – sin² θ)cos² θ

= cos² θ/ cos² θ

= 1

Given: (cosec θ – sin θ) = a³

(sec θ – cos θ) = b³

Since, a³ = (cosec θ – sin θ)

= (1/sin θ) – sin θ

=

= cos² θ/sin θ

Therefore, a² = (a³)^2/3 = (cos² θ/sin θ)^2/3

Also, b³ = sec θ – cos θ

= (1/cos θ) – cos θ

= (1 – cos² θ)/cos θ

= sin² θ/cos θ

Therefore, b² = (b³)^2/3 = (sin² θ/cos θ)^2/3

Now, consider the left – hand side:

a² b² (a² + b²) =

=

=

= sin² θ + cos² θ

= 1 = R.H.S.

Hence, proved.

Given: 2 sin θ + 3 cos θ = 2

Consider (2 sin θ + 3 cos θ)² + (3 sin θ – 2 cos θ)² = 4sin² θ + 9 cos² θ + 12sin θ cos θ + 9 sin² θ + 4 cos² θ – 12 sin θ cos θ

⇒ (2 sin θ + 3 cos θ)² + (3 sin θ – 2 cos θ)² = 13sin² θ + 13 cos² θ

⇒ (2 sin θ + 3 cos θ)² + (3 sin θ – 2 cos θ)² = 13(sin² θ + cos² θ)

⇒ (2 sin θ + 3 cos θ)² + (3 sin θ – 2 cos θ)² = 13

⇒ (2)² + (3 sin θ – 2 cos θ)² = 13

⇒ (3 sin θ – 2 cos θ)² = 13 – 4

⇒ (3 sin θ – 2 cos θ)² = 9

⇒ (3 sin θ – 2 cos θ) = ± 3

Hence, proved.

Given: (sin θ + cos θ) = √2 cos θ

To show: cot θ = (√2 + 1)

Solution:

(sin θ + cos θ) = √2 cos θ

Divide both sides by sin θ,







Since



⇒ 1 + cot θ = √2 cot θ

⇒ (√2 – 1)cot θ = 1

⇒ cot θ =

⇒ cot θ =

⇒ cot θ =

⇒ cot θ = √2 + 1

Given: cos θ + sin θ = √2sin θ

Consider (sin θ + cos θ)² + (sin θ – cos θ)² = sin² θ + cos² θ + 2sin θ cos θ + sin² θ + cos² θ –

2 sin θ cos θ

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = 2sin² θ + 2 cos² θ

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = 2(sin² θ + cos² θ)

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = 2

⇒ (√2sin θ)² + (sin θ – cos θ)² = 2

⇒ (sin θ – cos θ)² = 2 – 2sin² θ

⇒ (sin θ – cos θ)² = 2(1 – sin² θ)

⇒ (sin θ – cos θ)² = 2(cos² θ)

⇒ (sin θ – cos θ) = ± √2 cos θ

Hence, proved.

(i) Given: sec θ + tan θ = p ……(1)

Then, (sec θ + tan θ) × = p

⇒ = p

⇒ = p

⇒ sec θ – tan θ = (1/p) ……(2)

Adding equation (1) and (2), we get:

2 sec θ = p + (1/p)

⇒ sec θ =

Therefore, sec θ =

(ii) Given: sec θ + tan θ = p ……(1)

Then, (sec θ + tan θ) × = p

⇒ = p

⇒ = p

⇒ sec θ – tan θ = (1/p) ……(2)

Subtracting equation (2) from (1), we get:

2tan θ = p – (1/p)

⇒ tan θ =

(iii) Since sin θ = tan θ/sec θ

=

=

=

Given: tan A = n tan B

Therefore,

Thus,

Squaring both sides, we get,

⇒ cot² B = n²/tan² A ……(1)

Also, sin A = m sin B

Therefore, sin B = sin A/m

Thus, cosec B = m/sin A

⇒ cosec² B = m²/sin² A ……(2)

Now, subtract equation (2) from (1):

cosec² B – cot² B =

⇒ 1 =

⇒ 1 =

⇒ m² – n² cos² A = sin² A

⇒ m² – n² cos² A = 1 – cos² A

⇒ m² – 1 = n² cos² A – cos² A

⇒ (n² – 1)cos² A = m² – 1

⇒ cos² A = (m² – 1)/(n² – 1)

Hence, proved.

Given: m = (cos θ – sin θ)

n = (cos θ + sin θ)

Now, =

Multiply numerator and denominator by cos θ – sin θ :

Therefore, =

=

Now, =

Multiply numerator and denominator by cos θ + sin θ :

Therefore, =

=

Now, consider =

=

=

=

Divide numerator and denominator by cos θ:

=

=

Therefore, =

Hence, proved.

Consider (1 – sin²θ) sec² θ = (cos² θ) × sec² θ

(∵ sin² θ + cos² θ = 1)

= 1

Consider (1 – cos²θ) cosec² θ = (sin² θ) × cosec² θ

(∵ sin² θ + cos² θ = 1)

= 1

Consider (1 + tan²θ) cos² θ = (sec² θ) × cos² θ

(∵ 1 + tan² θ = sec² θ)

= 1

Consider (1 + cot²θ) × sin² θ = (cosec² θ) × sin² θ

(∵ 1 + cot² θ = cosec² θ)

= 1

Consider sin² θ +

= (sin²θ) + (1/sec² θ)

(∵ 1 + tan² θ = sec² θ)

= (sin²θ) + (cos²θ)

(∵ sin² θ + cos² θ = 1)

= 1

Consider cot² θ –

= (cot² θ) – (cosec² θ)

= – (cosec² θ – cot² θ)

(∵ 1 + cot² θ = cosec² θ)

= – 1

Consider sin θ cos (90° – θ) + cos θ sin (90° – θ) = sin θ sin θ + cos θ cos θ

= sin² θ + cos² θ

(∵ sin² θ + cos² θ = 1)

= 1

Consider cosec² (90° – θ) – tan² θ = sec² θ – tan² θ

(∵ 1 + tan² θ = sec² θ)

= 1

Consider sec² θ (1 + sin θ)(1 – sin θ) = sec² θ (1 – sin² θ)

= sec² θ cos² θ

(∵ sin² θ + cos² θ = 1)

= 1

Consider cosec² θ (1 + cos θ)(1 – cos θ) = cosec² θ (1 – cos² θ)

= cosec² θ sin² θ

(∵ sin² θ + cos² θ = 1)

= 1

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

= sin² θ cos² θ (sec² θ)(cosec² θ) (∵ 1 + cot² θ = cosec² θ and 1 + tan² θ = sec² θ)

= sin² θ (cosec² θ) cos² θ (sec² θ)

= 1 × 1

= 1

Consider (1 + tan² θ)(1 + sin θ)(1 – sin θ)

= (1 + tan² θ)(1 – sin² θ) (∵ sin² θ + cos² θ = 1 and 1 + tan² θ = sec² θ)

= (sec² θ) (cos² θ)

= 1

Consider 3 cot² θ – 3 cosec² θ = – 3(cosec² θ – cot² θ)

= – 3(1) (∵ 1 + cot² θ = cosec² θ)

= – 3

Consider 4 tan² θ – = 4 tan² θ – 4 sec² θ

= 4(tan² θ – sec² θ)

= 4( – 1) (∵1 + tan² θ = sec² θ)

= – 4

Consider = (∵1 + tan² θ = sec² θ and 1 + cot² θ = cosec² θ)

= 1

Give: sin θ = 1/2

Therefore cosec θ = 1/sin θ

= 2

Consider 3 cot² θ + 3 = 3 (cot² θ + 1)

= 3 cosec² θ (∵ 1 + cot² θ = cosec² θ)

= 3 (2)²

= 3 × 4

= 12

Give: cos θ = 2/3

Therefore sec θ = 1/cos θ

= 3/2

Consider 4 tan² θ + 4 = 4 (tan² θ + 1)

= 4 sec² θ (∵ 1 + tan² θ = sec² θ)

= 4 (3/2)²

= 4 × (9/4)

= 9

Given: cos θ = 7/25

Therefore sin θ = √(1 – cos² θ)

= √(1 –(49/625))

= √[(625 – 49)/625]

= √(576/625)

= 24/25

Thus, tan θ = sin θ/cos θ = (24/25)/(7/25)

= 24/7

Also, cot θ = 1/tan θ = 7/24

Therefore, tan θ + cot /θ = (24/7) + (7/24)

= (576 + 49)/(24 × 7)

= 625/168

Given: cos θ = 2/3

Thus, sec θ = 1/cos θ

= 3/2

Now, consider =

= [(1/2)/(5/2)]

= 1/5

Given: 5 tan θ = 4

Therefore, tan θ = 4/5

Now, consider and divide numerator and denominator by cos θ:

=

=

=

= (1/5)/(9/5)

= 1/9

Given: 3 cot θ = 4

Therefore, cot θ = 4/3

Therefore, tan θ = 3/4

Now, consider and divide numerator and denominator by cos θ:

=

=

=

= (11/4)/(13/4)

= 11/13

Given: cot θ = 1/√3

Thus, tan θ = 1/cot θ = √3

Therefore sec θ = √(1 + tan² θ)

= √(1 + 3)

= √(4)

= 2

Therefore sec² θ = 4

Now, cos² θ = 1/sec² θ = 1/4

So, consider =

=

=

= (3/4)/(5/4)

= 3/5

Given: tan θ = 1/√5

∴ tan² θ = 1/5

Consider =

Multiply numerator and denominator by sin θ:

=

=

= 4/6

= 2/3

We are given that: cot A = 4/3

⇒ tan (90° – A) = 4/3

Since A + B = 90°, therefore B = 90° – A

Therefore, tan (90° – A) = tan B = 4/3

We are given that: cos B = 3/5

⇒ sin (90° – B) = 3/5

Since A + B = 90°, therefore A = 90° – B

Therefore, sin (90° – B) = sin A = 3/5

We are given that: √3sin θ = cos θ

∴sin θ /cos θ = 1/√3

⇒ tan θ = 1/√3

⇒ tan θ = tan 30°

On comparing both sides, we get,

θ = 30°

Consider tan 10° tan 20° tan 70° tan 80°

= tan 10° tan 20° tan (90° – 20°) tan (90° – 10°)

= tan 10° tan 20° cot 0° cot 10°

= tan 10° cot 10° tan 20° cot 20°

= 1 × 1

= 1

Consider tan 1° tan 2°… tan 88° tan 89°

= tan 1° tan 2°… tan 44° tan 45° tan 46° …tan 88° tan 89°

= tan 1° tan 2°…tan 44° tan 45° tan (90° – 44° ) …tan (90° – 2°) tan (90° – 1°)

= tan 1° tan 2° … tan 44 ° tan 45° cot 44° …cot 2° cot 1°

= tan 1° cot 1° tan 2° cot 2°… tan 44 ° cot 44 ° tan 45°

= 1 × 1 × ...× 1

= 1

Consider cos 1° cos 2° cos 3° ... cos 180°

= cos 1° cos 2° cos 3° …× cos 90° × … × cos 180°

= cos 1° cos 2° cos 3° ... × 0 × … × cos 180°

= 0 (∵ cos 90° = 0)

Given: tan A = 5/12

Consider (sin A + cos A) sec A = (sin A + cos A)(1/cos A)

= (sin A/cos A) + (cos A/cos A)

= tan A + 1

= (5/12) + 1

= 17/12

We are given that: sin θ = cos (θ – 45°)

∴We can rewrite it as: cos (90° – θ ) = cos (θ – 45°)

On comparing both sides, we get,

90° – θ = θ – 45°

⇒ θ + θ = 90° + 45°

⇒ 2θ = 135°

⇒ θ = 65.5°

Consider – 4 cos 50° cosec 40°

= – 4 cos 50° cosec (90° – 40°)

= – 4 cos 50° sec 50°

= 1 + 1 – 4

= – 2

Consider sin 48° sec 42° + cos 48° cosec 42°

= sin 48° sec (90° – 48°) + cos 48° cosec (90° – 48°)

= sin 48° cosec 48° + cos 48° sec 48°

= 1 + 1

= 2

Given: x = a sin θ

y = b cos θ

Then b²x² + a²y² = b²(a sin θ)² + a²(b cos θ)²

= a²b² sin² θ + a² b² cos² θ

= a²b² (sin² θ + cos² θ)

= (a²b²) × 1

= a²b²

Given: 5x = sec θ, and 5/x = tan θ

Consider 5(x² – (1/ x²))

=

=

= (1/5) [sec² θ – tan² θ]

= (1/5)[1]

= 1/5 (∵ sec² x – tan² x = 1)

Given: 2x = cosec θ , and 2/x = cot θ

Consider 2(x² – (1/ x²)) =

=

= (1/2)(cosec² θ – cot² θ)

= 1/2 (∵ cosec² x – cot² x = 1)

Given: sec θ + tan θ = x ……(1)

Then, (sec θ + tan θ) × = x

⇒ = x

⇒ = x

⇒ sec θ – tan θ = (1/x) ……(2)

Adding equation (1) and (2), we get:

2 sec θ = x + (1/x)

= (x² + 1)/x

⇒ sec θ = (x² + 1)/2x

Therefore, sec θ = (x² + 1)/2x

Consider

=

=

=

=

= cot 60°

= 1/√3

Given: sin θ = x

Therefore, cosec θ = 1/x

Using the identity 1 + cot² θ = cosec² θ, we get

cot θ = √(cosec² θ – 1)

= √((1/x)² – 1)

=

=

Given: sec θ = x

Using the identity 1 + tan² θ = sec² θ, we get

tan θ = √(sec² θ – 1)

= √(x² – 1)

sec 30° = 1/cos 30°

= 1/(√3/2)

= 2/√3

cosec 60° = 1/sin 60°

= 1/(√3/2)

= 2/√3

Therefore, sec 30° /cosec 30° = (2/√3)/(2/√3)

= 1

=

=

= 1 + 1

= 2

Consider tan 10° tan 15° tan 75° tan 80°

= tan 10° tan 80° tan 15° tan 75°

= tan 10° tan (90 – 10)° tan 15° tan (90 – 15)°

= (tan 10° cot 10°) (tan 15° cot 15°)

= (1) × (1) = 1

Consider tan 5° tan 25° tan 30° tan 65° tan 85°

= tan 5° tan 85° tan 25° tan 65° tan 30°

= tan 5° tan (90 – 5)° tan 25° tan (90 – 25)° tan 30°

= (tan 5° cot 5°) (tan 25° cot 25°) tan 30°

= (1) × (1) × (1/√3)

= 1/√3

Consider cos 1° cos 2° cos 3° ... cos 180°

= cos 1° cos 2° cos 3° …× cos 90° × … cos 180°

= cos 1° cos 2° cos 3° ... × 0 × cos 180°

= 0 (∵ cos 90° = 0)

Consider =

=

=

= (2 + 1)/(3 – 1)

= 3

Consider (sin 47° cos 43°) + (cos 47° sin 43°)

= (sin 47° cos (90 – 47)°) + (cos 47° sin (90 – 47)°)

= sin² 47° + cos² 47°

= 1

Consider (sec 70° sin 20°) + (cos 20° cosec 70°)

= (sec (90 – 20)° sin 20°) + (cos 20° cosec (90 – 20)°)

= (cosec 70° sin70°) + (cos 20° sec 20°)

= 1 + 1 (∵ cosec θ = 1/sin θ and sec θ = 1/cos θ )

= 2

We are given that: sin 3A = cos (A – 10°)

∴ We can rewrite it as: cos (90° – 3A) = cos (A – 10°)

On comparing both sides, we get,

90° – 3A = A – 10°

⇒ A + 3A = 90° + 10°

⇒ 4A = 100°

⇒ A = 25°

We are given that: sec 4A = cosec (A – 10°)

∴We can rewrite it as: cosec (90° – 4A) = cosec (A – 10°)

On comparing both sides, we get,

90° – 4A = A – 10 °

⇒ A + 4A = 90° + 10°

⇒ 5A = 100°

⇒ A = 20°

We are given that: sin A = cos B

∴We can rewrite it as: sin A = sin(90° – B)

On comparing both sides, we get,

90° – B = A

⇒ A + B = 90°

We are given that: cos (α + β) = 0

∴We can rewrite it as: cos (α + β) = cos (90° – 0°)

On comparing both sides, we get,

(α + β) = 0 = 90°

⇒ α = 90° – β

Therefore, sin (α – β) = sin (90° – β – β)

= sin (90° – 2β)

= cos 2β

Consider sin (45° + θ) – cos (45° – θ) = sin (45° + θ) – sin (90° – (45° – θ))

= sin (45° + θ) – sin (45° + θ)

= 0

sec² 10° – cot² 80° = sec² 10° – tan² (90° – 80°)

= sec² 10° – tan² 10°

= 1

cosec² 57° – tan² 33° = cosec² 57° – cot² (90° – 33°)

= cosec² 57° – cot² 57°

= 1

Consider =

=

= (2tan² 30° × 1)/1

= 2tan² 30°

= 2(1/√3)²

= 2/3

Consider + sin² 63° + cos 63°sin 27°

= + sin² 63° + cos 63° sin (90° – 63°)

= + sin² 63° + cos 63° cos 63°

= (1/1) + (sin² 63° + cos² 63°)

= 1 + 1

= 2

Consider – (cos² 20° + cos² 70°)

= – (cos² 20° + cos² (90° – 70°))

= (tan θ × cot θ) + (tan 50°/tan 50°) – (cos² 20° + sin² 20°)

= 1 + 1 – 1

= 1

Consider

=

=

=

= 1/√3

Given: 2 sin 2θ = √3

Therefore, sin 2θ = √3/2

⇒ sin 2θ = sin 60°

On comparing both sides, we get:

2θ = 60°

⇒ θ = 60°/2

⇒ θ = 30°

Given: 2 cos 3θ = 1

Therefore, cos 3θ = 1/2

⇒ cos 3θ = cos 60°

On comparing both sides, we get:

3θ = 60°

⇒ θ = 60°/3

⇒ θ = 20°

Given√3tan 2θ – 3 = 0

Therefore, √3tan 2θ = 3

⇒ tan 2θ = 3/√3

⇒ tan 2θ = √3

⇒ tan 2θ = tan 60°

On comparing both sides, we get:

2θ = 60°

⇒ θ = 60°/2

⇒ θ = 30°

Given: tan x = 3 cot x

⇒ tan x/cot x = 3

Since, cot x = 1/tan x

Therefore, tan x/cot x = 3 ⇒ tan² x = 3

Taking square root on both sides:

⇒ tan x = √3

⇒ tan x = tan 60°

Comparing both sides:

⇒ x = 60°

Given: x tan 45° cos 60° = sin 60° cot 60°

⇒ x × 1 × (1/2) = (√3/2) × (1/√3)

⇒ x/2 = 1/2

⇒ x = 1

Given: tan²45° – cos²30° = x sin 45° cos 45°

⇒ (1)² – (√3/2)² = x × (1/√2) × (1/√2)

⇒ 1 – (3/4) = x × (1/2)

⇒ x/2 = 1 –(3/4)

⇒ x/2 = 1/4

⇒ x = 2/4

⇒ x = 1/2

sec² 60° – 1 = (1/cos² 60°) – 1

= [1/(1/2)²] – 1

= [1/(1/4)] – 1

= 4 – 1

= 3

Consider (cos 0° + sin 30° + sin 45°) (sin 90° + cos 60° – cos 45°)

= [1 + (1/2) + (1/√2)] × [1 + (1/2) – (1/√2)]

= [(3/2) + (1/√2)] × [(3/2) – (1/√2)]

= (3/2)² – (1/√2)²

= (9/4) – (1/2)

= (9 – 2)/4

= 7/4

Consider sin² 30° + 4 cot² 45° – sec² 60° = (1/2)² + 4(1)² – (2)²

= (1/4) + 4 – 4

= 1/4

Consider 3 cos²60° + 2 cot²30° – 5 sin²45° = 3 (1/2)² + 2(√3)² – 5(1/√2)²

= 3(1/4) + 2(3) – 5(1/2)

= (3/4) + 6 – (5/2)

= (3 + 24 – 10)/4

= 17/4

Consider cos² 30° cos² 45° + 4 sec² 60° + 2cos² 90° – 2 tan² 60°

Given: cosec θ = √10

Therefore, sin θ = 1/√10

Since, sin² θ + cos² θ = 1

Therefore, cos θ = √(1 – sin² θ)

=

=

=

=

Therefore, sec θ = 1/cos θ =

Given: tan θ = 8/15 = Perpendicular/Base

On comparing, we get:

Perpendicular = 8

Base = 15

Therefore, (Hypotenuse)² = (Perpendicular)² + (Base)²

= 64 + 225

= 289

Therefore, hypotenuse = √289

= 17

Therefore cosec θ = Hypotenuse/Perpendicular

= 17/8

Given: sin θ = a/b

Since, sin² θ + cos² θ = 1

Therefore, cos θ = √(1 – sin² θ)

=

=

Given: tan θ = √3 = Perpendicular/Base

On comparing, we get:

Perpendicular = √3

Base = 1

Therefore, (Hypotenuse)² = (Perpendicular)² + (Base)²

= 3 + 1

= 4

Therefore, hypotenuse = √4

= 2

Therefore sec θ = Hypotenuse/Base

= 2/1 = 2

Given: sec θ = 25/7

Therefore, cos θ = 1/sec θ = 7/25 = Base/Hypotenuse

Therefore, on comparing, Base = 7 and Hypotenuse = 25

In a right – angled triangle, (Hypotenuse)² = (Perpendicular)² + (Base)²

625 = (Perpendicular)² + 49

Therefore, Perpendicular = √(625 – 49) = √(576)

= 24

Therefore sin θ = Perpendicular/Hypotenuse

= 24/25

Given: sin θ = 1/2 = Perpendicular/Hypotenuse

Therefore, on comparing, Perpendicular = 1 and Hypotenuse = 2

In a right – angled triangle, (Hypotenuse)² = (Perpendicular)² + (Base)²

4 = 1+ (Base)²

Therefore, Base = √(4 – 1)

= √(3)

Therefore cot θ = Base/ Perpendicular

= √3/1

= √3

Given: cos θ = 4/5 = Base/Hypotenuse

Therefore, on comparing, Base = 4 and Hypotenuse = 5

In a right – angled triangle, (Hypotenuse)² = (Perpendicular)² + (Base)²

25 = (Perpendicular)² + 16

Therefore, Perpendicular = √(25 – 16) = √(9)

= 3

Therefore tan θ = Perpendicular/Base

= 3/4

Given: 3x = cosec θ , and 3/x = cot θ

Consider 3(x² – (1/ x²)) =

=

= (1/3)(cosec² θ – cot² θ)

= 1/3 (∵ cosec² x – cot² x = 1)

Given: 2x = sec A, and 2/x = tan A

Consider 2(x² – (1/ x²))

=

=

= (1/2) [sec² A – tan² A]

= (1/2)[1]

= 1/2 (∵ sec² x – tan² x = 1)

Given: tan θ = 4/3 = Perpendicular/Base

Therefore, (Hypotenuse)² = (Perpendicular)² + (Base)²

= 16 + 9

= 25

Therefore, hypotenuse = √25

= 5

Therefore sin θ = Perpendicular/Hypotenuse

= 4/5

Also, cos θ = Base/Hypotenuse

= 3/5

Thus, sin θ + cos θ = (4/5) + (3/5)

= 7/5

Given: tan θ + cot θ = 5

Squaring both sides, we get:

tan² θ + cot² θ + 2 tan θ cot θ = 25

tan² θ + cot² θ = 25 – 2 tan θ cot θ

= 25 – 2

= 23

Given: cos θ + sec θ = 5/2

Squaring both sides, we get:

cos² θ + sec² θ + 2 cos θ sec θ = 25/4

cos² θ + sec² = (25/4) – 2 cos θ sec θ

= 25/4 – 2

= (25 – 8)/4

= 17/4

Given: tan θ = 1/√7

∴ tan² θ = 1/7

Consider =

Multiply numerator and denominator by sin θ:

=

=

= 6/8

= 3/4

Given: 7tan θ = 4

Therefore tan θ = 4/7

Consider and divide numerator and denominator by cos θ:

=

=

=

= 1/7

Given: 3cot θ = 4

Therefore cot θ = 4/3

Consider and divide numerator and denominator by sin θ:

=

=

=

= 9

Given: tan θ = a/b

Consider and divide numerator and denominator by cos θ:

=

=

=

Given: sin A + sin² A = 1

Therefore sin A = 1 – sin² A = cos² A ……(1)

Now, consider cos²A + cos⁴A = cos² A(1 + cos²A)

Put the value of cos²A in the above equation:

Therefore, cos²A + cos⁴A = cos² A(1 + cos²A)

= (1 – sin² A)(1+1 – sin² A)

Again, from equation (1), we have 1 – sin² A = sin A. So, put the value of sin A in the above equation:

Therefore, cos²A + cos⁴A = (sinA)(1+ sinA)

= sin A + sin² A

= 1 (given)

Therefore, cos²A + scos⁴A = 1

Given: cos A + cos² A = 1

Therefore cos A = 1 – cos² A = sin² A ……(1)

Now, consider sin²A + sin⁴A = sin² A(1 + sin²A)

Put the value of sin²A in the above equation:

Therefore, sin²A + sin⁴A = sin² A(1 + sin²A)

= (1 – cos² A)(1+1 – cos² A)

Again, from equation (1), we have 1 – cos² A = cos A. So put the value of cos A in the above equation:

Therefore, sin²A + sin⁴A = (cosA)(1+ cosA)

= cos A + cos² A

= 1 (given)

Therefore, sin²A + sin⁴A = 1

Consider and rationalize:

=

=

=

=

=

= sec A – tan A

Consider and rationalize:

=

=

=

=

=

= cosec A + cot A

Given: tan θ = a/b

Consider and divide numerator and denominator by cos θ:

=

=

=

Consider (cosec θ – cot θ)² =

=

=

=

=

=

= R.H.S.

Hence, proved.

Consider (sec A + tan A)(1 – sin A) =

=

=

=

= cos A

Consider + 3 tan² 56° tan² 34°

= + 3 tan² 56° tan² (90° – 56°)

= + 3 tan² 56° cot² 56°

= (1/1) + 3(1)

= 1 + 3

= 4

Consider (sin² 30° cos² 45° + 4 tan² 30°+ (1/2) sin² 90^o + (1/8) cot² 60^o)

= [(1/2)² × (1/√2)² + 4 (1/√3)² + (1/2) × (1)² + (1/8)(1/√3)²]

= [(1/4) × (1/2)] + [(4/3)] + (1/2) + (1/24)

= (1/8) + (4/3) + (1/2) + (1/24)

= (3 + 32 + 12 + 1)/24

= 48/24

= 2

Given: cos A + cos² A = 1

Therefore cos A = 1 – cos² A = sin² A ……(1)

Now, consider sin²A + sin⁴A = sin² A(1 + sin²A)

Put the value of sin²A in the above equation:

Therefore, sin²A + sin⁴A = sin² A(1 + sin²A)

= (1 – cos² A)(1+1 – cos² A)

Again, from equation (1), we have 1 – cos² A = cos A. So put the value of cos A in the above equation:

Therefore, sin²A + sin⁴A = (cosA)(1+ cosA)

= cos A + cos² A

= 1 (given)

Therefore, sin²A + sin⁴A = 1

Given: sin θ = √3/2

Therefore, cosec θ = 1/sin θ = 2/√3

cos θ = √(1 – sin² θ)

= √(1 – (3/4))

= √(1/4)

= 1/2

cot θ = cos θ/sin θ = (1/2)/√3/2

= 1/√3

Therefore, (cosec θ + cot θ) = (2/√3) + (1/√3)

= 3/√3

= √3

Given: cot A = 4/5

Consider and divide numerator and denominator by sin A:

=

=

=

= 9

Given: 2x = sec A, and 2/x = tan A

Therefore,
(2x)² = sec²A

⇒ 4x² = sec²A [1]

and


[2]


On subtracting [2] from [1], we get

(∵ sec² x – tan² x = 1)

Given: √3 tanθ = 3 sinθ

⇒

⇒

⇒

⇒

Also, we know,

⇒

⇒

Now, we need to prove that:

sin²θ – cos²θ = 1/3

⇒ L.H.S =

⇒ L.H.S =

⇒ L.H.S = 1/3

Consider L.H.S. =

=

=

= 1/1

= 1

= R.H.S.

Hence, proved.

Given: 2 sin 2θ = √3

Therefore, sin 2θ = √3/2

⇒ sin 2θ = sin 60°

⇒ 2θ = 60°/2 = 30°

Therefore, θ = 30°

Consider and rationalize:

=

=

=

=

=

= cosec A + cot A

Given: cosec θ + cot θ = p

p² – 1 = (cosec θ + cot θ)² – 1

= cosec² θ + cot² θ + 2 cosec θ cot θ – 1

= cosec² θ – 1 + cot² θ + 2 cosec θ cot θ

= cot² θ + cot² θ + 2 cosec θ cot θ

= 2 cot θ (cot θ + cosec θ)

Also, p² + 1 = (cosec θ + cot θ)² + 1

= cosec² θ + cot² θ + 2 cosec θ cot θ + 1

= cosec² θ + 1 + cot² θ + 2 cosec θ cot θ

= cosec² θ + cosec² θ + 2 cosec θ cot θ

= 2 cosec θ (cosec θ + cot θ)

Now, consider L.H.S. =

=

= cot θ/cosec θ

= cos θ

= R.H.S.

Hence, proved.

Consider R.H.S. = and rationalize:

=

=

=

=

=

= (cosec A – cot A)²

= L.H.S.

Hence, proved.

Given: 5cot θ = 3

Therefore cot θ = 3/5

Consider and divide numerator and denominator by sin θ:

=

=

=

= 16/29

Hence, showed.

Consider L.H.S. = (sin 32° cos 58° + cos 32° sin 58°)

= sin 32° cos (90° – 32°) + cos 32° sin (90° – 32°)

= sin² 32° + cos² 32°

= 1

= R.H.S.

Hence, proved.

Given: a sin θ + b cos θ = x …….(1)

a cos θ – b sin θ = y …….(2)

Square equation (1) and (2) on both sides:

a² sin² θ + b² cos² θ + 2ab cos θ sin θ = x² …….(3)

a² cos² θ + b² sin² θ – 2ab cos θ sin θ = y² ……..(4)

Add equation (3) and (4):

[a² sin² θ + b² cos² θ + 2ab cos θ sin θ] + [a² cos² θ + b² sin² θ – 2ab cos θ sin θ] = x² + y²

⇒ a² (sin² θ + cos² θ) + b² (cos² θ + sin² θ) = x² +y²

⇒ a² +b² = x² + y²

Hence, proved.

Consider L.H.S. =

Multiply numerator and denominator by (1 + sin θ):

=

=

=

=

= [(1/cos θ) + (sin θ/cos θ)]²

= (sec θ + tan θ)²

= R.H.S.

Hence, proved.

Consider L.H.S. =

Multiply and divide the first term by (sec θ + tan θ):

=

= – sec θ

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

= tan θ

Consider R.H.S. =

Multiply and divide the second term by (sec θ – tan θ):

=

= sec θ –

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

= tan θ

Therefore, L.H.S. = R.H.S.

Hence, proved.

Consider L.H.S. =

=

=

= sin A/cos A

= tan A

= R.H.S.

Hence, proved.

Consider L.H.S. =

=

=

=

=

=

=

=

=

= tan A + (1/tan A) + 1

= 1 + tan A + cot A

= R.H.S.

Hence, proved.

We are given that: sec 5A = cosec (A – 36°)

∴We can rewrite it as: cosec (90° – 5A) = cosec (A – 36°)

On comparing both sides, we get,

90° – 5A = A – 36°

⇒ A + 5A = 90° + 36°

⇒ 6A = 126°

⇒ A = 21°

Hence, proved.