Question

Pls say answer for this immediately

Answer 1

$\large \bf\underline{\underline {\green{Solution:-}}}\\\\\sf \small We\;need \;to\;prove:\\\\\sf\sqrt{\dfrac{1-cos\theta}{1+cos\theta}}=cosec\theta - cot\theta \\\\\rm Taking\;LHS\\\\\rm By \; rationalizing\;the\; denominator\\\\\rm To\;rationalize\;we\;have\;to\;multiply\;\&\;divide\;with\;(1-cos\theta)\\\\\longrightarrow\sf\sqrt{\dfrac{(1-cos\theta)(1-cos\theta)}{(1+cos\theta)(1-cos\theta)}}$

Using

• $\tt (a-b)(a-b)=(a-b)^2$
• $\tt (a+b)(a-b)=a^2 - b^2$

$\sf\longrightarrow \sqrt{\dfrac{(1-cos\theta)^2}{1^2-cos^2\theta}}\\\\\sf\longrightarrow \dfrac{\sqrt{(1-cos\theta)^2}}{\sqrt{sin^2\theta}}\;\;\;\;\;\;(\because 1-cos^2\theta=sin^2\theta)\\\\\sf\longrightarrow \dfrac{1-cos\theta}{sin\theta}\\\\\sf\longrightarrow\dfrac{1}{sin\theta} -\dfrac{cos\theta}{sin\theta}\\\\\dag \;\;\large \underline{\boxed{\purple{\bf cosec\theta-cos\theta}}}$

By comparing with RHS

$\leadsto\bf cosec\theta- cot\theta=cosec\theta-cot\theta$

LHS = RHS

Hence , proved!

$\large\underline {\underline{\green{\bf More \; Information:-}}}$

$\boxed{\begin{minipage}{7cm}\bf \underline{</p><p>Important trigonometric identities}\\</p><p>\sf \underline{Identity 1:} sin^2\theta+cos^2\theta=1\\\sf\underline{Identity 2:} sec^2\theta -tan^2\theta=1\\\sf \underline{Identity 3:}cosec^2\theta -cot^2\theta \end{minipage}}$