Question

(cosec theta - sin theta)( sec theta- cos theta)= 1/ tan theta +cos theta
prove this ​

Answer 1

$\bold\blue{Correct \ Question}$

$\bold{(cosec \ \theta-sin \ \theta)(sec \ \theta-cos \ \theta)=sin \ \theta \ . \ cos \ theta}$

$\bold{prove \ this}$

$\bold\red{\underline{\underline{Answer:}}}$

$\bold\pink{To \ prove:}$

$\bold{(cosec \ \theta-sin \ \theta)(sec \ \theta-cos \ \theta)=sin \ \theta \ . \ cos \ \theta}$

$\bold\green{\underline{\underline{Proof:}}}$

$\bold{L.H.S.=(cosec \ \theta-sin \ \theta)(sec \ \theta-cos \ \theta)}$

$\bold{(cosec=\frac{1}{sin})}$

$\bold{(sec=\frac{1}{cos})}$

$\bold{Trigonometric \ ratio}$

$\bold{=(\frac{1}{sin \ \theta}-sin \ \theta)(\frac{1}{cos \ \theta}-cos \ \theta)}$

$\bold{=(\frac{1-sin^{2}\theta}{sin \ \theta})(\frac{1-cos^{2}\theta}{cos \ \theta})}$

$\bold{(1-sin^{2}\theta=cos^{2}\theta)}$

$\bold{1-cos^{2}\theta=sin^{2}\theta}$

$\bold{... Trigonometric \ identity}$

$\bold{=(\frac{cos^{2}\theta}{sin \ \theta})(\frac{sin^{2}\theta}{cos \ theta})}$

$\bold{=(cos \ \theta \ . \ cot \ \theta)(sin \ \theta \ . \ tan \ \theta}$

$\bold{cot=\frac{1}{tan}}$

$\bold{Trigonometric \ ratio}$

$\bold{=(cos \ \theta \ . \ \frac{1}{tan \ \theta})(sin \ theta \ . \ tan \ \theta)}$

$\bold{=sin \ \theta.cos \ \theta}$

$\bold{=R.H.S}$

$\bold{Hence \ proved}$

$\bold\purple{\tt{(cosec \ \theta-sin \ \theta)(sec \ \theta-cos \ \theta)=sin \ \theta \ . \ cos \ \theta}}$