Question

if tan theta =a/b prove that cos theta + sin theta/ cos theta - sin theta =b+a/b-a​

Answer 1

GIVEN :-)

$\tan\theta = \dfrac{a}{b}$

TO PROVE :-)

$\dfrac{cos \theta + sin \theta }{cos \theta - sin \theta } = \dfrac{b + a}{b - a}$

LHS:-)

$\dfrac{cos \theta + sin \theta }{cos \theta - sin \theta }$

Dividing Numerator And Denominator by

Cosθ we get

$\dfrac{1 + tan \theta }{1 - tan \theta} \\ \\ = \dfrac{1 + \dfrac{a}{b} }{1 - \dfrac{a}{b} } \\ \\ = \dfrac{ \dfrac{b + a}{ \cancel{b}} }{ \dfrac{b - a}{ \cancel{b}} } \\ \\ = \dfrac{b + a}{b - a} = RHS$

HENCE PROVED.

Answer 2

Here is your Solution ☺☺☺✌✌✌

LHS

$\dfrac{cos \theta + sin \theta }{cos \theta - sin \theta }$

Dividing Numerator And Denominator by

Cosθ. we get

$\dfrac{1 + tan \theta }{1 - tan \theta} \\ \\ = \dfrac{1 + \dfrac{a}{b} }{1 - \dfrac{a}{b} } \\ \\ = \dfrac{ \dfrac{b + a}{ {b}} }{ \dfrac{b - a}{ {b}} } \\ \\ = \dfrac{b + a}{b - a} = RHS$

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