Assertion (A): If A=15^{circ}, B=17^{circ} an | vedclass

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(A) Reason: In $	riangle PQR$,$P+Q+R=180^{\circ}$.$\frac{P}{2} + \frac{Q}{2} + \frac{R}{2} = 90^{\circ} \Rightarrow \frac{P}{2} + \frac{Q}{2} = 90^{\

Vedclass · Accepted Answer

$(A)$ is true,$(R)$ is true and $(R)$ is the correct explanation for $(A)$

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(A) Reason: In $	riangle PQR$,$P+Q+R=180^{\circ}$.$\frac{P}{2} + \frac{Q}{2} + \frac{R}{2} = 90^{\circ} \Rightarrow \frac{P}{2} + \frac{Q}{2} = 90^{\circ} - \frac{R}{2}$.Taking tangent on both sides: $	an(\frac{P}{2} + \frac{Q}{2}) = 	an(90^{\circ} - \frac{R}{2}) = \cot \frac{R}{2}$.$\frac{	an \frac{P}{2} + 	an \frac{Q}{2}}{1 - 	an \frac{P}{2} 	an \frac{Q}{2}} = \frac{1}{	an \frac{R}{2}}$.$(	an \frac{P}{2} + 	an \frac{Q}{2}) 	an \frac{R}{2} = 1 - 	an \frac{P}{2} 	an \frac{Q}{2}$.$	an \frac{P}{2} 	an \frac{Q}{2} + 	an \frac{Q}{2} 	an \frac{R}{2} + 	an \frac{R}{2} 	an \frac{P}{2} = 1$. Thus,$(R)$ is true.Assertion: Given $A=15^{\circ}, B=17^{\circ}, C=13^{\circ}$,then $2A+2B+2C = 2(15^{\circ}+17^{\circ}+13^{\circ}) = 2(45^{\circ}) = 90^{\circ}$.Let $P=4A, Q=4B, R=4C$. Then $P+Q+R = 4(15^{\circ}+17^{\circ}+13^{\circ}) = 180^{\circ}$.Using the identity from $(R)$ with $P=4A, Q=4B, R=4C$:$	an 2A 	an 2B + 	an 2B 	an 2C + 	an 2C 	an 2A = 1$.Substituting $	an 	heta = \frac{1}{\cot 	heta}$:$\frac{1}{\cot 2A \cot 2B} + \frac{1}{\cot 2B \cot 2C} + \frac{1}{\cot 2C \cot 2A} = 1$.$\frac{\cot 2C + \cot 2A + \cot 2B}{\cot 2A \cot 2B \cot 2C} = 1$.$\cot 2A + \cot 2B + \cot 2C = \cot 2A \cot 2B \cot 2C$. Thus,$(A)$ is true and $(R)$ is the correct explanation.

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