જો alpha, beta લઘુકોણ હોય કે જેથી sin beta=2 | vedclass

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Question

(A) આપેલ છે: $\sin \beta=2 \sin \alpha \dots (i)$ અને $3 \cos \beta=2 \cos \alpha \Rightarrow \cos \beta=\frac{2}{3} \cos \alpha \dots (ii)$ સમીકરણોનો

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) આપેલ છે: $\sin \beta=2 \sin \alpha \dots (i)$ અને $3 \cos \beta=2 \cos \alpha \Rightarrow \cos \beta=\frac{2}{3} \cos \alpha \dots (ii)$ સમીકરણોનો વર્ગ કરીને સરવાળો કરતા: $\sin^2 \beta + \cos^2 \beta = (2 \sin \alpha)^2 + (\frac{2}{3} \cos \alpha)^2$ $1 = 4 \sin^2 \alpha + \frac{4}{9} \cos^2 \alpha$ $1 = 4 \sin^2 \alpha + \frac{4}{9} (1 - \sin^2 \alpha)$ $1 = 4 \sin^2 \alpha + \frac{4}{9} - \frac{4}{9} \sin^2 \alpha$ $1 - \frac{4}{9} = \frac{32}{9} \sin^2 \alpha$ $\Rightarrow \frac{5}{9} = \frac{32}{9} \sin^2 \alpha$ $\Rightarrow \sin^2 \alpha = \frac{5}{32}$ તેવી જ રીતે,$\cos^2 \alpha = 1 - \frac{5}{32} = \frac{27}{32}$ હવે,$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$ $\cos \beta = \frac{2}{3} \cos \alpha$ અને $\sin \beta = 2 \sin \alpha$ મુકતા: $\cos(\alpha + \beta) = \cos \alpha (\frac{2}{3} \cos \alpha) - \sin \alpha (2 \sin \alpha)$ $\cos(\alpha + \beta) = \frac{2}{3} \cos^2 \alpha - 2 \sin^2 \alpha$ $\cos(\alpha + \beta) = \frac{2}{3} (\frac{27}{32}) - 2 (\frac{5}{32}) = \frac{18}{32} - \frac{10}{32} = \frac{8}{32} = \frac{1}{4}$ તેથી,$\sec(\alpha + \beta) = \frac{1}{\cos(\alpha + \beta)} = 4$.

List-$I$	List-$II$
$(I)$ $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \sin^2 15^{\circ} + \dots + \sin^2 90^{\circ}$	$(A)$ $0$
$(II)$ $\tan^2 5^{\circ} \cdot \tan^2 10^{\circ} \cdot \tan^2 15^{\circ} \dots \tan^2 85^{\circ}$	$(B)$ $\frac{19}{2}$
$(III)$ $\cos^2 5^{\circ} + \cos^2 10^{\circ} + \cos^2 15^{\circ} + \dots + \cos^2 180^{\circ}$	$(C)$ $18$
$(IV)$ $\cot 5^{\circ} + \cot 10^{\circ} + \cot 15^{\circ} + \dots + \cot 175^{\circ}$	$(D)$ $1$
	$(E)$ $-1$

જો $\alpha, \beta$ લઘુકોણ હોય કે જેથી $\sin \beta=2 \sin \alpha$ અને $3 \cos \beta=2 \cos \alpha$ થાય,તો $\sec (\alpha+\beta)=$

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