$\sin 70=\cos \theta,$ then $\theta=\ldots \ldots \ldots \ldots$
$70$
$90$
$20$
$30$
$\sin 70=\sin (90-20)=\cos 20 .$
But, $\sin 70=\cos \theta . $
$ \therefore \theta=20$
$\sin (90-\theta)=\ldots \ldots \ldots$
$\cos \theta=\frac{b}{\sqrt{a^{2}+b^{2}}} ;$ where, $0<\theta<90 ;$ then $\sin \theta=\ldots \ldots \ldots \ldots$
Write 'True' or 'False' and justify your answer.
$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}=1$
$\sin 48 \sec 42+\cos 48 \operatorname{cosec} 42=\ldots \ldots \ldots \ldots$
$8 \sin ^{2} 45-2 \tan ^{2} 60+3 \cot ^{2} 30-2 \cos ^{2} 45$
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