Question

prove--(cos A / cosec A+1)+ ( cos a /cosec A -1)=2tan A​

Answer 1

Answer:

Step-by-step explanation:

:Given:

$\tt \dfrac{cos\: A}{cosec\:A+1} + \dfrac{cos\:A}{cosec\:A-1} =2\:tan\:A$

To Prove:

LHS = RHS

Proof:

Taking the LHS of the equation,

$\tt \dfrac{cos\: A}{cosec\:A+1} + \dfrac{cos\:A}{cosec\:A-1}$

Taking cos A outside,

$\tt \implies cos\:A \times \bigg(\dfrac{1}{cosec\:A+1} + \dfrac{1}{cosec\:A-1}\bigg)$

Cross multiplying we get,

$\tt \implies cos\:A \bigg( \dfrac{cosec\:A-1+cosec\:A+1}{(cosec\:A+1)\times (cosec\:A-1)} \bigg)$

We know,

(a + b) × (a - b) = a² - b²

$\tt \implies cos\:A \bigg( \dfrac{2\:cosec\:A}{cosec^{2} \:A-1} \bigg)$

We know that,

cosec²A - 1 = cot² A

Substituting it,

$\tt \implies cos\:A \bigg( \dfrac{2\:cosec\:A}{cot^{2} \:A} \bigg)$

$\tt \implies 2\times cosec\:A \times cos\:A \times \dfrac{1}{cot^{2}\:A }$

We know,

cosec A = 1/sin A

cot A = cos A/sin A

Hence,

$\tt \implies 2\times \dfrac{1}{sin\:A} \times cos\:A \times \dfrac{1}{\dfrac{cos^{2}\:A}{sin^{2}\:A } }$

$\tt \implies 2\times \dfrac{1}{sin\:A} \times cos\:A \times \dfrac{sin^{2}\:A }{cos^{2}\:A }$

$\tt \implies 2\times \dfrac{sin\:A}{cos\:A}$

We know that,

sin A/cos A = tan A

$\tt \implies 2\:tan\:A$

= RHS

Hence proved,