In a triangle ABC if tan A2 - p and tan B2 - q- prove that sin C - 2p - q1 - pq1 - p21 - q2

We have R.H.S = 2(p+q)(1 pq)(1+p^2)(1+q^2) 2[tan(A/2) + tan(B/2)][1 tan(A/2) tan(B/2)][1+tan^2(A/2)][1+tan(B/2)] 2sin 12(A+B) cos 12(A+B) (sin^2 1 ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios of Compound Angles

In a triangle...

Question

In a triangle ABC if $tan$ $\frac{A}{2}$ = p and tan $\frac{B}{2}$ = q, prove that $sin C = \frac{2 (p + q) (1 - p q)}{(1 + p^{2}) (1 + q^{2})}$

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Solution

We have R.H.S = $\frac{2 (p + q) (1 - p q)}{(1 + p^{2}) (1 + q^{2})}$
$\frac{2 [t a n (A / 2) + t a n (B / 2)] [1 - t a n (A / 2) t a n (B / 2)]}{[1 + t a n^{2} (A / 2)] [1 + t a n (B / 2)]}$
$\frac{2 s i n \frac{1}{2} (A + B) c o s \frac{1}{2} (A + B)}{(s i n^{2} \frac{1}{2} A + c o s^{2} \frac{1}{2} A) (s i n^{2} \frac{1}{2} B + c o s^{2} \frac{1}{2} B)}$
$= 2 s i n \frac{1}{2} (A + B) c o s \frac{1}{2} (A + B)$
$= s i n (A + B) = s i n (π - C) = s i n C$

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In a triangle ABC if tan A2 = p and tan B2 = q, prove that sinC=2(p+q)(1−pq)(1+p2)(1+q2)

We have R.H.S =2(p+q)(1−pq)(1+p2)(1+q2)2[tan(A/2)+tan(B/2)][1−tan(A/2)tan(B/2)][1+tan2(A/2)][1+tan(B/2)]2sin12(A+B)cos12(A+B)(sin212A+cos212A)(sin212B+cos212B)=2sin12(A+B)cos12(A+B)=sin(A+B)=sin(π−C)=sinC

In a triangle ABC if $tan$ $\frac{A}{2}$ = p and tan $\frac{B}{2}$ = q, prove that $sin C = \frac{2 (p + q) (1 - p q)}{(1 + p^{2}) (1 + q^{2})}$