Dear Student, 3 tan θ=4tan θ=43LHS3 sin θ + 2 cos θ3 sin θ-2 cos θ=cos θ 3 sin θcos θ + 2cos θ 3 sin θcos θ-2=3 tan θ+23 tan θ-2=343+2343-2=62=3=RHSHence Proved Regards

Pls. Solve Qn.No. 14

Dear Student,

3 \tan θ = 4 \tan θ = \frac{4}{3} LHS \frac{3 \sin θ + 2 \cos θ}{3 \sin θ - 2 \cos θ} = \frac{\cos θ (3 \frac{\sin θ}{\cos θ} + 2)}{\cos θ (3 \frac{\sin θ}{\cos θ} - 2)} = \frac{3 \tan θ + 2}{3 \tan θ - 2} = \frac{3 (\frac{4}{3}) + 2}{3 (\frac{4}{3}) - 2} = \frac{6}{2} = 3 = RHS Hence Proved .

Regards

-2

View Full Answer