Question

Prove that: (secθ+tanθ-1)/(tanθ-secθ+1)= cosθ/(1-sinθ)
KINDLY ANSWER ACCORDING TO CLASS 10 MATHS

Answer 1

TO PROVE THAT :-

$\sf \dfrac{sec\theta+tan\theta- 1}{tan\theta-sec\theta+1}= \dfrac{cos\theta}{1-sin\theta}$

SOLUTION :-

$\sf L.H.S = \dfrac{sec\theta+tan\theta-1}{tan\theta-sec\theta+1}$

         $= \sf \dfrac{sec\theta+tan\theta-(sec^2\theta-tan^2\theta)}{tan\theta-sec\theta+1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ sec^2\theta-tan^2\theta= 1 \ ]\\\\= \dfrac{sec\theta+tan\theta-(sec\theta+tan\theta)\times (sec\theta-tan\theta)}{tan\theta-sec\theta+1}$

Take $\sf sec\theta+tan\theta$ as common,

        $= \sf \dfrac{sec\theta+tan\theta \ [1-(sec\theta-tan\that)]}{tan\theta- sec\theta+1}\\\\= \dfrac{sec\theta+tan\theta \ (1-sec\theta+tan\theta)}{tan\theta-sec\theta+1}\\\\= \dfrac{sec\theta+tan\theta \ (tan\theta- sec\theta+1)}{tan\theta-sec\theta+1} \\\\= sec\theta+tan\theta\\\\= \dfrac{1}{cos\theta}+\dfrac{sin \theta}{cos\theta}\\\\= \dfrac{1+sin\theta}{cos\theta}$

Multiplying both numerator and denominator by $\sf ( 1-sin\theta)$ .

        $= \sf \dfrac{(1+sin^2\theta)(1-sin\theta)}{cos\theta(1-sin\theta)}\\\\= \dfrac{1-sin^2\theta}{cos\theta(1-sin\theta)}\\\\= \dfrac{cos^2\theta}{\cos\theta(1-sin\theta)}\\\\= \dfrac{cos\theta}{1-sin\theta} \\\\= R.H.S$

Hence proved.