Question

what will be the value of thera in:-
4sin^2 theta+ 8cos^2 theta=4​

Answer 1

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

Given:

• $4\sin^2 \theta+ 8\cos^2 \theta=4$

To find:

• $Value\:of\:\theta$

Solution:

$\implies 4\sin^2 \theta+ 8\cos^2 \theta=4 \\ \implies 4(\sin^2 \theta + 2\cos^2 \theta) = 4 \\ \implies \sin^2 \theta + 2\cos^2 \theta = 1 \\ \implies \sin^2 \theta + \cos^2 \theta + \cos^2 \theta = 1 \\ \implies 1 + \cos^2 \theta = 1 \\ \implies \cos^2 \theta = 0 \\ \\ \implies \theta = \dfrac{\pi}{2}^c = 90^{\circ}. \\ \\ This\:is\:bcoz\:\cos 90^{\circ}=0.$

Formula Used:

• $\sin^2 \theta + \cos^2 \theta = 1$

Learn More:

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