Question

please solve it plz ​

Answer 1

Answer:

R.H.S , Hence proved in the above attachment.

hope it helps you!

Answer 2

To Prove :-

$\sf (cosec \theta - cot \theta)^2 = \dfrac{1- cos \theta}{1+cos \theta}$

Solution :-

$\sf LHS=(cosec \theta - cot \theta)^2$

$\dag \bf \ cosec \theta = \dfrac{1}{sin \theta}$

$\dag \bf \ cot \theta = \dfrac{cos \theta}{sin \theta}$

$\sf:\implies\left( \dfrac{1}{sin \theta} - \dfrac{cos\theta}{sin \theta} \right)^2$

$:\implies\sf \dfrac{(1-cos \theta)^2}{sin^2 \theta}$

$\dag\bf \ sin^2 \theta = 1- cos^2 \theta$

$\sf:\implies\dfrac{(1-cos \theta)^2}{1-cos^2 \theta}$

$\sf :\implies \dfrac{(1-cos\theta)(1-cos\theta)}{(1+cos\theta)(1-cos\theta)}$

$\bf :\implies \dfrac{1-cos\theta}{1+cos\theta} =RHS$

Hence proved!

★ Trigonometric Identities :

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$