Trigonometry> trigono (e1.21)...

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???

Aditi Bhatnagar , 14 Years ago

Grade 12th Pass

1 Answers

Jitender Singh

Ans:Hello student, please find answer to your question
$tanB = cotP.tanA$
$tanP = \frac{tanA}{tanB}$
$tanP = \frac{2tan\frac{P}{2}}{1-tan^{2}\frac{P}{2}}$
$\frac{2tan\frac{P}{2}}{1-tan^{2}\frac{P}{2}} = \frac{tanA}{tanB}$
$tan^{2}\frac{P}{2} + 2\frac{tanB}{tanA}.tan\frac{P}{2}-1 = 0$
$tan\frac{P}{2} = \frac{-2\frac{tanB}{tanA}\pm \sqrt{4\frac{tan^{2}B}{tan^{2}A}+4}}{2}$
$tan\frac{P}{2} = -\frac{tanB}{tanA}\pm \sqrt{\frac{tan^{2}B}{tan^{2}A}+1}$
$tan\frac{P}{2} = \frac{-tanB\pm \sqrt{tan^{2}B+tan^{2}A}}{tanA}$
$tan^{2}\frac{P}{2} = \frac{tan^{2}B+tan^{2}B+tan^{2}A\mp 2tanB.\sqrt{tan^{2}A +tan^{2}B}}{tan^{2}A}$
$cot^{2}\frac{P}{2} = \frac{tan^{2}A}{2tan^{2}B+tan^{2}A\mp 2tanB.\sqrt{tan^{2}A +tan^{2}B}}$