Question

Prove the following :
$\sin ({50}^{\circ }+\theta )-\cos ({40}^{\circ }-\theta )+\tan 1^{\circ }\tan {10}^{\circ }\tan {20}^{\circ }\tan {70}^{\circ }\tan {80}^{\circ }\tan {89}^{\circ }=1$

Answer 1

L.H.S$=\sin ({50}^{\circ }+\theta )-\cos ({40}^{\circ }-\theta )+\tan 1^{\circ }\tan {10}^{\circ }\tan {20}^{\circ }\tan {70}^{\circ }\tan {80}^{\circ }\tan {89}^{\circ }$

We know that,

$\cos (90-\theta )=\sin \theta$

$\sin (90-\theta )=\cos \theta$

$\tan (90-\theta )=\cot \theta$

$=\sin ({90}^{\circ }-({40}^{\circ }-\theta \left)\right)-\cos ({40}^{\circ }-\theta )+$

$\tan ({90}^{\circ }-{89}^{\circ })\tan ({90}^{\circ }-{80}^{\circ })\tan ({90}^{\circ }-{70}^{\circ })\tan {70}^{\circ }\tan {80}^{\circ }\tan {89}^{\circ }$

$=\cos ({40}^{\circ }-\theta )-\cos ({40}^{\circ }-\theta )+\cot {89}^{\circ }\cot {80}^{\circ }\cot {70}^{\circ }\tan {70}^{\circ }\tan {80}^{\circ }\tan {89}^{\circ }$

We have $\cot \theta =\frac{1}{\tan \theta }$

$=\cot {89}^{\circ }\cot {80}^{\circ }\cot {70}^{\circ }\tan {70}^{\circ }\tan {80}^{\circ }\tan {89}^{\circ }$

$=\frac{\tan {70}^{\circ }\tan {80}^{\circ }\tan {89}^{\circ }}{\tan {70}^{\circ }\tan {80}^{\circ }\tan {89}^{\circ }}$

$=1$=R.H.S

$\therefore \sin ({50}^{\circ }+\theta )-\cos ({40}^{\circ }-\theta )+\tan 1^{\circ }\tan {10}^{\circ }\tan {20}^{\circ }\tan {70}^{\circ }\tan {80}^{\circ }\tan {89}^{\circ }=1$

hence proved