कथन (A): यदि alpha=12^{circ}, beta=15^{circ}, | vedclass

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(A) For Assertion $(A)$: Given $\alpha=12^{\circ}, \beta=15^{\circ}, \gamma=18^{\circ}$.We check if $	an 2 \alpha 	an 2 \beta+	an 2 \beta 	an 2 \g

Vedclass · Accepted Answer

$(A)$ और $(R)$ दोनों सत्य हैं और $(R)$,$(A)$ का सही स्पष्टीकरण है

Vedclass · Answer

(A) For Assertion $(A)$: Given $\alpha=12^{\circ}, \beta=15^{\circ}, \gamma=18^{\circ}$.We check if $	an 2 \alpha 	an 2 \beta+	an 2 \beta 	an 2 \gamma+	an 2 \gamma 	an 2 \alpha=1$.This is equivalent to $	an 2 \alpha (	an 2 \beta + 	an 2 \gamma) = 1 - 	an 2 \beta 	an 2 \gamma$.$	an 2 \alpha = \frac{1 - 	an 2 \beta 	an 2 \gamma}{	an 2 \beta + 	an 2 \gamma} = \frac{1}{	an(2 \beta + 2 \gamma)} = \cot(2 \beta + 2 \gamma)$.Since $2 \alpha + 2 \beta + 2 \gamma = 2(12^{\circ} + 15^{\circ} + 18^{\circ}) = 2(45^{\circ}) = 90^{\circ}$,we have $2 \alpha = 90^{\circ} - (2 \beta + 2 \gamma)$.Thus,$	an 2 \alpha = 	an(90^{\circ} - (2 \beta + 2 \gamma)) = \cot(2 \beta + 2 \gamma)$.So,Assertion $(A)$ is true.For Reason $(R)$: In $	riangle ABC$,$A+B+C = 180^{\circ}$,so $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^{\circ}$.Then $\frac{A}{2} + \frac{C}{2} = 90^{\circ} - \frac{B}{2}$.Taking tangent on both sides: $	an(\frac{A}{2} + \frac{C}{2}) = 	an(90^{\circ} - \frac{B}{2}) = \cot \frac{B}{2}$.$\frac{	an \frac{A}{2} + 	an \frac{C}{2}}{1 - 	an \frac{A}{2} 	an \frac{C}{2}} = \frac{1}{	an \frac{B}{2}}$.$	an \frac{B}{2} (	an \frac{A}{2} + 	an \frac{C}{2}) = 1 - 	an \frac{A}{2} 	an \frac{C}{2}$.$	an \frac{A}{2} 	an \frac{B}{2} + 	an \frac{B}{2} 	an \frac{C}{2} + 	an \frac{C}{2} 	an \frac{A}{2} = 1$.Reason $(R)$ is true and it provides the general identity that explains the specific case in $(A)$.

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