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

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(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}}$.

$1. \cos \theta$	$a. \frac{\cos \theta}{\sin \theta}$
$2. \tan \theta$	$b. \frac{1}{\csc \theta}$
$3. \cot \theta$	$c. \frac{1}{\sec \theta}$
$4. \sin \theta$	$d. \frac{1}{\cot \theta}$
	$e. \sin \theta \cdot \cos \theta$

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

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