Question

prove that :cos7x+cos5x/sin7x+sin5x=cotx​

Answer 1

Required Answer:-

Given to prove:

$\rm \mapsto \dfrac{ \cos(7x) + \cos(5x) }{ \sin(7x) - \sin(5x) } = \cot(x)$

Proof:

We know that,

$\rm \mapsto \cos( \alpha ) + \cos( \beta ) = 2 \cos \bigg(\dfrac{ \alpha + \beta }{2} \bigg) \cos \bigg( \dfrac{ \alpha - \beta }{2} \bigg)$

$\rm \mapsto \sin( \alpha ) - \sin( \beta ) = 2 \cos \bigg( \dfrac{ \alpha + \beta }{2} \bigg) \sin \bigg( \dfrac{ \alpha - \beta }{2} \bigg)$

Taking LHS,

$\rm\dfrac{ \cos(7x) + \cos(5x) }{ \sin(7x) - \sin(5x) }$

$\rm = \dfrac{2 \cos \bigg( \dfrac{7x + 5x}{2}\bigg) \cos \bigg( \dfrac{7x - 5x}{2} \bigg) }{2 \cos \bigg( \dfrac{7x + 5y}{2} \bigg) \sin \bigg( \dfrac{7x - 5x}{2} \bigg) }$

$\rm = \dfrac{ \cancel{2} \cancel{\cos(6x)}\cos (x)}{ \cancel{2} \cancel{\cos(6x)}\sin(x)}$

$\rm = \dfrac{ \cos(x) }{ \sin(x) }$

$\rm = \dfrac{1}{ \tan(x) }$

$\rm = \cot(x)$

Taking RHS,

$\rm = \cot(x)$

Hence, LHS = RHS (Proved)

Relationship between different Trigo Functions:

1. sin(x) = 1/cosec(x)
2. cos(x) = 1/sec(x)
3. tan(x) = 1/cot(x)
4. sin(x)/cos(x) = tan(x)