Question

if 2tan theta = 3, then sin^2theta + cos^2theta =?

Answer 1

Answer:

answer = 1 hope u got your answer

Answer 2

Given :-

$\sf\bullet \ 2 \ tan \theta = 3$

$\sf\therefore tan \theta = \dfrac{3}{2}$

To Find :-

$\sf sin^2 \theta+cos^2 \theta$

Solution :-

We know :-

$\bf\dag \ sec^2 \theta = 1+ tan^2 \theta$

$\sf : \implies sec^2 \theta = 1 + \left( \dfrac{3}{2}^2 \right)$

$\sf : \implies sec^2 \theta = 1+ \dfrac{9}{4}$

$\sf : \implies sec^2 \theta = \dfrac{4+9}{4}$

$\sf : \implies sec^2 \theta = \dfrac{13}{4}$

We know :-

$\bf\dag \ cos \theta = \dfrac{1}{sec \theta }$

$\sf : \implies cos^2 \theta = \dfrac{1}{sec^2 \theta }$

$\sf : \implies cos^2 \theta = \dfrac{4}{13}$

We know :-

$\bf\dag \ sin \theta = cos \theta tan \theta$

$\sf : \implies sin^2 \theta = cos^2 \theta tan^2 \theta$

$\sf : \implies sin^2 \theta = \dfrac{4}{13} \times \dfrac{9}{4}$

$\sf : \implies sin^2 \theta = \dfrac{9}{13}$

$\sf : \implies sin^2 \theta + cos^2 \theta = \dfrac{9}{13}+\dfrac{4}{13}$

$\sf : \implies sin^2 \theta+cos^2 \theta = \dfrac{9+4}{13}$

$\sf : \implies sin^2 \theta + cos^2 \theta = \dfrac{13}{13}$

$\bf : \implies sin^2 \theta +cos^2 \theta = 1$