Question

Solve ques 3,4 and 5 only....please solve it's urgent!!!!

Answer 1

Trigonometry......... After a long tym...... Here is ur solution

Answer 2

(iii) sinθsin(90-θ) - cosθcos(90-θ)
   = sinθcosθ - cosθsinθ
   = 0
(iv) sec(90-θ)cosec(90-θ)
     =cosecθsecθ
     =(1/sinθ)(1/cosθ)
     =(1/sinθ)(cosθ/cos²θ)
     =$\frac{cos}{sinθ cos^{2}θ }$
     =$\frac{cosθ}{sinθ} * \frac{1}{cos²θ}$
     =cotθ* sec²θ

(v) $\frac{sin(90-θ)}{cosec(90-θ)} + \frac{cos(90-θ)}{sec(90-θ)} \\= \frac{cosθ}{secθ} + \frac{sinθ}{cosecθ} \\= (cosθ* \frac{1}{secθ})+(sinθ* \frac{1}{cosecθ}) \\ = (cosθ*cosθ)+(sinθ*sinθ) \\ = cos^{2}θ+sin^{2}θ \\ =1$