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If tan B= cot P tan A, then cot 2 (P/2) is equal to (a) sin(A+B) / sin (A-B) (B) sin(A-B) / sin (A+B) (c) cos(A+B) / cos (A-B) (D) cos(A-B) / cos (A+B) Explain pls how???

If tan B= cot P tan A, then  cot2(P/2) is equal to


(a) sin(A+B) / sin (A-B)  (B) sin(A-B) / sin (A+B)
(c) cos(A+B) / cos (A-B)  (D) cos(A-B) / cos (A+B)


Explain pls how???

Grade:12th Pass

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:Hello student, please find answer to your question
tanB = cotP.tanA
tanP = \frac{tanA}{tanB}
tanP = \frac{2tan\frac{P}{2}}{1-tan^{2}\frac{P}{2}}
\frac{2tan\frac{P}{2}}{1-tan^{2}\frac{P}{2}} = \frac{tanA}{tanB}
tan^{2}\frac{P}{2} + 2\frac{tanB}{tanA}.tan\frac{P}{2}-1 = 0
tan\frac{P}{2} = \frac{-2\frac{tanB}{tanA}\pm \sqrt{4\frac{tan^{2}B}{tan^{2}A}+4}}{2}
tan\frac{P}{2} = -\frac{tanB}{tanA}\pm \sqrt{\frac{tan^{2}B}{tan^{2}A}+1}
tan\frac{P}{2} = \frac{-tanB\pm \sqrt{tan^{2}B+tan^{2}A}}{tanA}
tan^{2}\frac{P}{2} = \frac{tan^{2}B+tan^{2}B+tan^{2}A\mp 2tanB.\sqrt{tan^{2}A +tan^{2}B}}{tan^{2}A}
cot^{2}\frac{P}{2} = \frac{tan^{2}A}{2tan^{2}B+tan^{2}A\mp 2tanB.\sqrt{tan^{2}A +tan^{2}B}}

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