Question

if a is not in second quadrant and b is not in3rd quadrant and cosa=-1/2 and cosB = -1/2 then 4sinB -3tana/ tanB + sin a=?​

Answer 1

$i) It \: is \: given \: that \: \angle A \:is \:not \:in$

$Q_{2} \: So \: it \: lies \: in \: Q_{3}$

$In \: \triangle OQR ,$

$Cos A = \frac{-1}{2}$

/* By Phythagorean theorem */

$OQ^{2} + QR^{2} = OR^{2}$

$\implies (-1)^{2} + QR^{2} = 2^{2}$

$\implies 1 + QR^{2} = 4$

$\implies QR^{2} = 4 - 1$

$\implies QR = \sqrt{3}\: --(1)$

$ii) It \: is \: given \: that \: \angle B \:is \:not \:in$

$Q_{3} \: So \: it \: lies \: in \: Q_{2}$

$In \: \triangle OQP ,$

$Cos B = \frac{-1}{2}$

/* By Phythagorean theorem */

$OQ^{2} + QP^{2} = OP^{2}$

$\implies (-1)^{2} + QP^{2} = 2^{2}$

$\implies 1 + QP^{2} = 4$

$\implies QP^{2} = 4 - 1$

$\implies QP = \sqrt{3}\: --(2)$

$Now, sin B = \frac{PQ}{OP} = \frac{\sqrt{3}}{2}$

$tan A = \frac{QR}{OQ} = \frac{\sqrt{3}}{(-1)}$

$tan B = \frac{PQ}{OQ} = \frac{\sqrt{3}}{(-1)}$

$Now, sin A = \frac{QR}{OR} = \frac{\sqrt{3}}{2}$

$\red{ Value \: of \: 4Sin B - \frac{3 tan A}{tan B} + sin A }$

$= 4 \times \frac{\sqrt{3}}{2} - 3\Big( \frac{\frac{\sqrt{3}}{(-1)}}{\frac{\sqrt{3}}{(-1)}}\Big) + \frac{\sqrt{3}}{2}$

$= 2\sqrt{3} - 3 + \frac{\sqrt{3}}{2}$

$= \frac{ 4\sqrt{3} - 6 + \sqrt{3}}{2}$

$= \frac{5\sqrt{3} - 6}{2}$

Therefore.,

$\red{ Value \: of \: 4Sin B - \frac{3 tan A}{tan B} + sin A }$

$\green{=\frac{5\sqrt{3} - 6}{2}}$

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