Prove that:(i) sinθ cos(90°-θ)+sin(90°-θ) cosθ=1(ii) sinθcos(90°-θ)+cosθsin(90°-θ)=2(iii) sinθ cos(90°-θ)cosθsin(90°-θ)+cosθ sin(90°-θ)sinθcos(90°-θ)=1(iv) cos(90°-θ)sec(90°-θ)tanθcosec(90°-θ)sin(90°-θ)cot(90°-θ)+tan(90°-θ)cotθ=2(v) cos(90°-θ)1+sin(90°-θ)+1+sin(90°-θ)cos(90°-θ)=2cosecθ(vi) sec90°-θ cosecθ-tan90°-θ cotθ+cos225°+cos265°3tan27° tan63°=23 CBSE 2010(vii) cotθ tan90°-θ-sec90°-θcosecθ+3tan12° tan60° tan78°=2 CBSE 2010

Prove that:(i) sinθ cos(90°-θ)+sin(90°-θ) cosθ=1(ii) sinθcos(90

1.	Prove that:(i) sinθ cos(90°-θ)+sin(90°-θ) cosθ=1(ii) sinθcos(90°-θ)+cosθsin(90°-θ)=2(iii) sinθ cos(90°-θ)cosθsin(90°-θ)+cosθ sin(90°-θ)sinθcos(90°-θ)=1(iv) cos(90°-θ)sec(90°-θ)tanθcosec(90°-θ)sin(90°-θ)cot(90°-θ)+tan(90°-θ)cotθ=2(v) cos(90°-θ)1+sin(90°-θ)+1+sin(90°-θ)cos(90°-θ)=2cosecθ(vi) sec90°-θ cosecθ-tan90°-θ cotθ+cos225°+cos265°3tan27° tan63°=23 CBSE 2010(vii) cotθ tan90°-θ-sec90°-θcosecθ+3tan12° tan60° tan78°=2 CBSE 2010
Answer» Prove that: (i) $sinθ \cos (90 ° - θ) + \sin (90 ° - θ) cosθ = 1$ (ii) $\frac{sinθ}{\cos (90 ° - θ)} + \frac{cosθ}{\sin (90 ° - θ)} = 2$ (iii) $\frac{sinθ \cos (90 ° - θ) cosθ}{\sin (90 ° - θ)} + \frac{cosθ \sin (90 ° - θ) sinθ}{\cos (90 ° - θ)} = 1$ (iv) $\frac{\cos (90 ° - θ) \sec (90 ° - θ) tanθ}{cosec (90 ° - θ) \sin (90 ° - θ) \cot (90 ° - θ)} + \frac{\tan (90 ° - θ)}{cotθ} = 2$ (v) $\frac{\cos (90 ° - θ)}{1 + \sin (90 ° - θ)} + \frac{1 + \sin (90 ° - θ)}{\cos (90 ° - θ)} = 2 cosecθ$ (vi) $\frac{\sec (90 ° - θ) cosecθ - \tan (90 ° - θ) cotθ + \cos^{2} 25 ° + \cos^{2} 65 °}{3 \tan 27 ° \tan 63 °} = \frac{2}{3} [CBSE 2010]$ (vii) $cotθ \tan (90 ° - θ) - \sec (90 ° - θ) cosecθ + \sqrt{3} \tan 12 ° \tan 60 ° \tan 78 ° = 2 [CBSE 2010]$