Question

prove that. (sec A + tan A) (1- sin A) = cos A​

Answer 1

Prove !

(secA + tanA ) (1- sinA) = cos A

$\sf \bullet sec A= \dfrac{1}{cosA}\\ \\ \sf \bullet tan\ A= \dfrac{sinA}{cosA} \\ \\ :\implies\sf \bigg(\dfrac{1}{cosA}+\dfrac{sinA}{cosA}\bigg)\times \bigg(1-sinA\bigg)\\ \\ :\implies\sf \dfrac{(1+sinA)\times (1-sinA)}{cosA}\\ \\ \sf \bullet \ It\ is \ in \ form \ of :-\\ \\ \sf \ \ a^2-b^2= (a+b)(a-b) \\ \\ :\implies\sf \dfrac{(1-sin^2A)}{cosA}\\ \\ \bullet \sf \ 1-sin^2A= cos^2A\\ \\ :\implies \sf \cancel{\dfrac{cos^2A}{cosA}}\\ \\ :\implies \sf cosA \\ \\ :\implies\sf cosA= cosA \\ \\ \sf \ \ L. H.S = R.H.S \ \ (hence \ proved \ !)$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

${\star{\sf{\green{ (secA + tanA)(1 - sinA) = cosA}}}}$

${\sf{\underline{\green{Now,}}}}$

Take L.H.S,

$\diamond{\bf{\blue{we \: know \: that}}}$

$\boxed{\tt{secA = \frac{1}{cosA} }}$

$\boxed{\tt{tanA = \frac{sinA}{cosA} }}$

$\implies{\tt{ \bigg( \frac{1}{cosA} + \frac{sinA}{cosA} \bigg) (1 - sinA)}}$

$\implies{\tt{ \bigg( \frac{1 + sinA}{cosA} \bigg) (1 - sinA)}}$

$\implies{\tt{ \bigg( \frac{1}{cosA} \bigg) (1 + sinA)(1 - sinA)}}$

$\implies{\tt{ \bigg( \frac{1}{cosA} \bigg) ( {1}^{2} - sinA^{2} )}}$

$\because\boxed{\tt{ {a}^{2} - {b}^{2} = (a - b)(a + b) }}$

$\implies{\tt{ \bigg( \frac{1}{cosA} \bigg) ( cosA^{2} )}}$

$\implies{\tt{ cosA }}$

$\boxed{\tt{ \purple{Hence \: L.H.S=R.H.S}}}$