Question

Please answer the following question

Answer 1

taking LHS, :

sin A/ sec A +1+ sin A/ sec A-1

= sin A ( sec A-1 )+ sin A ( sec A +1 )/ sec^2 A - 1

=2sin A . sec A / tan^ A

=2 sin A × 1 / cos A × 1/ tan^2A

=2 tanA × 1 / tan^2A

=2/ tan A  = 2 cot A

therefore LHS = RHS, hence proved

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Answer 2

To prove ----->

SinA / ( SecA + 1 ) + SinA / ( SecA - 1 ) = 2 CotA

Proof ------->

LHS = SinA / ( SecA + 1 ) + SinA / ( SecA - 1 )

Taking SinA common from numerator , we get,

= SinA { 1 / ( SecA + 1 ) + 1 / ( SecA - 1 ) }

= SinA{(SecA - 1) + (SecA + 1)/(SecA - 1) (SecA + 1 )}

We know that , a² - b² = ( a + b ) ( a - b ) , applying it we get,

= SinA { 2 SecA / ( SecA )² - ( 1 )² }

= SinA { 2SecA /( Sec²A - 1 ) }

We know that , Sec²θ - 1 = tanθ , applying it we get,

= SinA ( 2 SecA / tan²A )

= 2 SinA SecA ( 1 / tan²A )

We know that, Secθ = 1 / Cosθ , applying it we get,

= 2 SinA ( 1 / CosA ) ( 1 / tan²A )

= 2 ( SinA / CosA ) ( 1 / tan²A )

We know that , tanθ = Sinθ / Cosθ , we get,

= 2 tanA ( 1 / tan²A )

= 2 ( 1 / tanA )

We know that , Cotθ = 1 / tanθ , we get,

= 2 CotA = R H S