cos theta = frac{b}{sqrt{a^2 + b^2}}; where,0 | vedclass

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Question

(A) We know the trigonometric identity: $\sin^2 	heta + \cos^2 	heta = 1$.Therefore,$\sin^2 	heta = 1 - \cos^2 	heta$.Substituting the given value

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$\frac{a}{\sqrt{a^2 + b^2}}$

Vedclass · Answer

(A) We know the trigonometric identity: $\sin^2 	heta + \cos^2 	heta = 1$.Therefore,$\sin^2 	heta = 1 - \cos^2 	heta$.Substituting the given value of $\cos 	heta = \frac{b}{\sqrt{a^2 + b^2}}$:$\sin^2 	heta = 1 - \left( \frac{b}{\sqrt{a^2 + b^2}} ight)^2 = 1 - \frac{b^2}{a^2 + b^2}$.Taking the common denominator:$\sin^2 	heta = \frac{(a^2 + b^2) - b^2}{a^2 + b^2} = \frac{a^2}{a^2 + b^2}$.Since $0 Thus,$\sin 	heta = \sqrt{\frac{a^2}{a^2 + b^2}} = \frac{a}{\sqrt{a^2 + b^2}}$.

Part $I$	Part $II$
$1.$ $\cos(90^\circ - \theta)$	$a.$ $\sec \theta$
$2.$ $\cot(90^\circ - \theta)$	$b.$ $\sin \theta$
$3.$ $\operatorname{cosec}(90^\circ - \theta)$	$c.$ $1$
	$d.$ $\tan \theta$

$\cos \theta = \frac{b}{\sqrt{a^2 + b^2}}$; where,$0 < \theta < 90^\circ$; then $\sin \theta = \dots$

Similar Questions

Which of the following groups truly matches the data of Part $I$ with the data of Part $II$?

Part $I$ Part $II$

$1.$ $\cos(90^\circ - \theta)$ $a.$ $\sec \theta$

$2.$ $\cot(90^\circ - \theta)$ $b.$ $\sin \theta$

$3.$ $\operatorname{cosec}(90^\circ - \theta)$ $c.$ $1$

$d.$ $\tan \theta$

Prove that,$(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)=\sec \alpha+\operatorname{cosec} \alpha$

If $\sec ^{2} \theta+\tan ^{2} \theta=\frac{13}{12}$,then the value of $\sec ^{4} \theta-\tan ^{4} \theta$ is .........

$\tan ^{2} \theta - \sec ^{2} \theta = \ldots \ldots \ldots$

Prove that $(\sin^{4} \theta - \cos^{4} \theta + 1) \operatorname{cosec}^{2} \theta = 2$.

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