Question

(sinθ+cosecθ)² + (cosθ+secθ)² = 7 + tan²θ+cot²θ,Prove it by using trigonometric identities.

Answer 1

Given :

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

L.H.S - (sinθ+cosecθ)² + (cosθ+secθ)²

= sin²θ+cosec²θ + 2sinθ.cosecθ + cos²θ+sec²θ + 2cosθsecθ

[ (a+b)² = a² + b² + 2ab]

= (sin²θ + cos²θ) + 2sinθ.cosecθ + 2cosθsecθ + cosec²θ + sec²θ

= 1 + 2sinθ.1/sinθ + 2cosθ . 1/cosθ + cosec²θ + sec²θ

[cosecθ = 1/sinθ , secθ = 1/cosθ]

= 1 + 2 + 2 + cosec²θ + sec²θ

= 5 + (1+ cot²θ) + (1 + tan²θ)

[ cosec²θ=1+ cot²θ , sec²θ = 1 + tan²θ]

= 5 + 1+ 1+ cot²θ + tan²θ

L.H.S = 7 + tan²θ + cot²θ = R.H.S

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

Here I am using A instead of

theta .

LHS = ( sinA + cosecA)²

+ ( CosA + secA )²

= (Sin²A+cosec²A+2sinAcosecA)

+ (Cos²A+sec²A+2cosAsecA)

______________________

( a + b )² = a² + b² + 2ab

______________________

= Sin²A+cosec² A + 2

+ Cos²A+sec²A + 2

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Since ,

i ) sinA × cosecA = 1

ii ) cosA × secA = 1

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Rearranging the terms , we get

= ( Sin²A + cos²A ) + cosec²A

+ Sec²A + 4

= ( Sin²A + cos²A )+( 1 + cot²A )

+ ( 1 + tan²A ) + 4

*******************************

By trigonometric identities:

i ) sin²A + cos²A = 1

ii ) cosec²A = 1 + cot² A

iii ) sec²A = 1 + tan² A

***********************************

= 1 + 1 + cot²A + 1 + tan² A + 4

= 7 + tan²A + cot² A

= RHS

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