Question

Evaluate:

$\sin ({50}^{\circ }+\theta )-\cos ({40}^{\circ }-\theta )+\tan 1^{\circ }\tan {15}^{\circ }\tan {20}^{\circ }\tan {70}^{\circ }\tan {65}^{\circ }\tan {89}^{\circ }+\sec ({90}^{\circ }-\theta ).cosec\theta -\tan ({90}^{\circ }-\theta )\cot \theta$

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

2

$\sin (50°+\theta )-\cos (40°-\theta )+\tan 1°\tan 15°\tan 20°\tan 70°\tan 75°\tan 89°+\sec (90°-\theta ).cosec-tan({90}^o-\theta )cot\theta =$

$\cos (90°-(50°+\theta ))-\cos (40°-\theta )+\cot (90°-15°).\cot (90°-20°)\tan 75°\tan 89°+cosec\theta .cosec\theta -\cot \theta .\cot \theta$

$=\cos (40°-\theta )-\cos (40°-\theta )+\cot 89°\cot 75°\cot 70°\tan 70°\tan 75°\tan 89°+cose{c}^2\theta -{\cot }^2\theta$

$=1+1[\therefore {cosec}^2\theta -{\cot }^2\theta =1]$

$=2$

Hence, the answer is $2.$