Question

Please answer this question. ​

Answer 1

${\huge{\red{\sf{Given}}}}\begin{cases}\leadsto\bf{\dfrac{1+cos\theta}{sin\theta}-\dfrac{sin\theta}{1+cos\theta}=2cot\theta} \end{cases}$

${\huge{\red{\sf{To\:Prove}}}}\begin{cases}\leadsto \bf{LHS=RHS} \end{cases}$

$\huge\red{\underline{\bf{\green{Answer}}}}$

$\sf{\red{\longmapsto Taking \:the\:given\: equation}}$

$\sf{\purple{\hookrightarrow \dfrac{1+cos\theta}{sin\theta}-\dfrac{sin\theta}{1+cos\theta}=2cot\theta}}$

$\sf{\pink{\longrightarrow Taking\:LHS}}$

$\sf{\longmapsto\dfrac{1+cos\theta}{sin\theta}-\dfrac{sin\theta}{1+cos\theta}}$

${\underline{\sf{\orange{\leadsto Taking\:the\:LCM}}}}$

$\sf{\longmapsto\dfrac{(1+cos\theta)^{2}-sin^{2}\theta}{sin\theta(1+cos\theta)}}$

$\sf{\longmapsto\dfrac{1+cos^{2}\theta+2cos\theta-sin^{2}\theta}{sin\theta(1+cos\theta)}}$

$\large\pink{\boxed{\purple{\bf{(a+b)^{2}=a^{2}+b^{2}+2ab}}}}$

$\sf{\longmapsto\dfrac{1+cos^{2}\theta+2cos\theta-(1-cos^{2}\theta)}{sin\theta(1+cos\theta)}}$

$\sf{\longmapsto\dfrac{1+cos^{2}\theta+2cos\theta-1+cos^{2}\theta}{sin\theta(1+cos\theta)}}$

$\sf{\longmapsto \dfrac{2cos^{2}\theta+2cos\theta}{sin\theta(1+cos\theta)}}$

$\sf{\longmapsto 2cos\theta\dfrac{1+cos\theta}{sin\theta(1+cos\theta)}}$

$\sf{\longmapsto 2cos\theta\dfrac{\cancel{1+cos\theta}}{sin\theta(\cancel{1+cos\theta})}}$

$\sf{\longmapsto\dfrac{2cos\theta}{sin\theta}}$

$\sf{\longmapsto 2(\dfrac{cos\theta}{sin\theta})}$

$\sf{\pink{\longmapsto 2cot\theta}}$

${\underline{\sf{\orange{=RHS}}}}$

${\underline{\red{\bf{Hence\:Proved}}}}$