Question

Please answer with clear and proper solution.

★ Q - 30 PLEASE ★

THANK U IN ADVANCE.☺

Answer 1

Hola there,

${ (tan \alpha + cosec \beta )}^{2} - {(cot \beta - sec \alpha )}^{2} = 2tan \alpha cot \beta (cosec \alpha + sec \beta )$

LHS

=> ${(tan \alpha + cosec \beta )}^{2} - {(cot \beta - sec \alpha )}^{2}$

=> $( {tan}^{2} \alpha + {cosec}^{2} \beta + 2tan \alpha cosec \beta ) - ( {cot}^{2} \beta + {sec}^{2} \alpha - 2cot \beta sec \alpha )$

=> $({tan}^{2} \alpha - {sec}^{2} \alpha )+ ({cosec}^{2} \beta - {cot}^{2} \beta )+ 2tan \alpha cosec \beta + 2cot \beta sec \alpha$

=> $- 1 + 1 + 2 \times \frac{sin \alpha }{cos \alpha } \times \frac{1}{ sin \beta } + 2 \times \frac{cos \beta }{sin \beta } \times \frac{1}{cos \alpha }$

=> $2 \frac{(sin \alpha + cos \beta )}{cos \alpha sin \beta }$

RHS

=> $2tan \alpha cot \beta (cosec \alpha + sec \beta )$

=> $2 \frac{sin \alpha }{cos \alpha } \frac{cos \beta }{sin \beta } ( \frac{1}{sin \alpha } + \frac{1}{cos \beta } )$

=> $2 \frac{sin \alpha cos \beta }{cos \alpha sin \beta } ( \frac{cos \beta + sin \alpha }{sin \alpha cos \beta } )$

=> $2(\frac{sin \alpha + cos \beta }{cos \alpha sin \beta } )$

Therefore,

LHS = RHS

Hence Proved.

Hope this helps...:)

Answer 2

(tanα + cosecβ)² - (cotβ - secα)² = 2tanαcotβ(cosecα+secβ)

Opening The Brackets 

tan²α + cosec²β + 2tanαcosecβ - (cot²β + sec²α -2cotαsecβ) = RHS

tan²α + cosec²β + 2 tanαcosecβ - cot²β - sec²α + 2cotαsecβ = RHS

tan²α - sec²α + cosec²β - cot²β + 2 tanαcosecβ + 2cotαsecβ = RHS

secα²- 1 - secα² + cot²β + 1 - cot²β + 2 tanαcosecβ + 2cotαsecβ = RHS

On Solving We Will Get...

2tanαcosecβ + 2cotαsecβ = RHS

2 sinα/cosα (1 / sinβ) + 2 cosβ/sinβ  (1 /cosα) = RHS

2(sinα + cosβ)/cosα sinβ =  2 tanα cotβ (cosecα + secβ)
2(sinα + cosβ)/cosα sinβ =  2sinα/cosβ x cosβ/sinβ (1/sinα + 1/cosβ)

So, 2(sinα + cosβ)/cosα sinβ = 2(sinα + cosβ)/cosα sinβ 

RHS = LHS ......... HENCE PROVED