Question

$(\sin\:\theta\:+\:\cosec\:\theta)^{2}\:+\:(\cos\: \theta\:+\:\sec\: \theta)^{2}\:=\:7+{\tan}^{2}\: \theta\:+\:{cot}^{2}\:\theta$
Please guys help fast. Today is my exam.​

Answer 1

GIVEN:

• (sin θ + cosec θ)² + (cos θ + sec θ) = 7 + tan² θ + cot² θ

TO PROVE:

• (sin θ + cosec θ)² + (cos θ + sec θ) = 7 + tan² θ + cot² θ

FORMUALE USED:

★ (A + B)² = A² + 2AB + B²

★ Sin θ = 1 / Cosec θ

★ Cos θ = 1 / Sec θ

★ Sin² θ + Cos² θ = 1

★ Sec² θ = 1 + Tan² θ

★Cosec² θ = 1 + Cot² θ

PROOF:

Take (sin θ + cosec θ)² + (cos θ + sec θ)² as L.H.S and 7 + tan² θ + cot² θ as R.H.S

★L.H.S:

(sin θ + cosec θ)² + (cos θ + sec θ)²

(A + B)² = A² + 2AB + B²

(sin² θ + cosec² θ + 2sin θ cosec θ + cos² θ + sec² θ + 2 sec θ cos θ)

Sin θ = 1 / Cosec θ

Cos θ = 1 / Sec θ

sin² θ + cos² θ + cosec² θ + 2 (cosec θ / cosec θ) + sec² θ + 2 (sec θ / sec θ )

Sin² θ + Cos² θ = 1

1 + cosec² θ + 2 + sec² θ + 2

5 + cosec² θ + sec² θ

Sec² θ = 1 + Tan² θ

Cosec² θ = 1 + Cot² θ

5 + 1 + tan² θ + 1 + cot² θ

7 + tan² θ + cot² θ

★R.H.S

7 + tan² θ + cot² θ

L.H.S = R.H.S

(sin θ + cosec θ)² + (cos θ + sec θ) = 7 + tan² θ + cot² θ

★★HENCE PROVED★★

Answer 2

Need to prove:

(sinθ + cosecθ)² + (cosθ + secθ)² = 7 + tan²θ + cot²θ

Solution:

Firstly taking LHS from the given question

(sinθ + cosecθ)² + (cosθ + secθ)²

Simplifying using (a + b)² = a² + 2ab + b²

→ sin²θ + cosec²θ + 2(sinθ)(cosecθ) + cos²θ + sec²θ + 2(cosθ)(secθ)

Here , sinθcosecθ = 1 because sinθ & cosθ both are reciprocal to each other . Similarly cosθsecθ = 1

→ (sin²θ + cos²θ) + 2 + 2 + cosec²θ + sec²θ

→ 1 + 2 + 2 + cosec²θ + sec²θ. [°.° 1st identity]

Now , substituting

• cosec²θ = 1 + cot²θ. [°.° 3rd identity]

• sec²θ = 1 + tan²θ. [°.° 2nd identity]

→ 5 + 1 + cot²θ + 1 + tan²θ

→ 7 + cot²θ + tan²θ. [LHS]

Now , comparing with RHS

7 + cot²θ + tan²θ = 7 + cot²θ + tan²θ

LHS = RHS

Hence , proved !