Question
Verify that the equation $‘sin2 \theta + cos2 \theta = 1’$ is true when $\theta = 0^\circ$ or $\theta = 90^\circ.$
✓
Answer
$sin^2 \theta + cos^2 \theta = 1$
i. lf $\theta = 0^\circ,$
LH.S. $= sin^2 \theta + cos^2 \theta$
$= sin^2 0^\circ + cos^2 0^\circ$
$= 0 + 1 …[\because sin 0^\circ = 0, cos 0^\circ = 1]$
$= R.H.S.$
$\therefore sin^2 \theta + cos^2 \theta = 1ii. If \theta = 90^\circ,$
L.H.S.=$ sin^2 \theta +cos^2 \theta$
$= sin^2 90^\circ + cos^2 90^\circ$
$= 1 + 0 … [ \because sin 90^\circ = 1, cos 90^\circ = 0]$
$= 1$
= R.H.S.
$\therefore sin^2 \theta + cos^2 \theta = 1$
i. lf $\theta = 0^\circ,$
LH.S. $= sin^2 \theta + cos^2 \theta$
$= sin^2 0^\circ + cos^2 0^\circ$
$= 0 + 1 …[\because sin 0^\circ = 0, cos 0^\circ = 1]$
$= R.H.S.$
$\therefore sin^2 \theta + cos^2 \theta = 1ii. If \theta = 90^\circ,$
L.H.S.=$ sin^2 \theta +cos^2 \theta$
$= sin^2 90^\circ + cos^2 90^\circ$
$= 1 + 0 … [ \because sin 90^\circ = 1, cos 90^\circ = 0]$
$= 1$
= R.H.S.
$\therefore sin^2 \theta + cos^2 \theta = 1$
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