Quantitative Aptitude Ques 1979

Question: What is $ \sin 25{}^\circ \sin 35{}^\circ \sec 65{}^\circ \sec 55{}^\circ $ equal to?

Options:

A) $ -1 $

B) $ 0 $

C) $ \frac{1}{2} $

D) $ 1 $

Show Answer

Answer:

Correct Answer: D

Solution:

  • $ \sin 25{}^\circ \sin 35{}^\circ \sec 65{}^\circ \sec 55{}^\circ $ $ =\sin 25{}^\circ \sin 35{}^\circ .\frac{1}{\cos 65{}^\circ }.\frac{1}{\cos 55{}^\circ } $ $ =\sin 25{}^\circ .\sin 35{}^\circ \times \frac{1}{\cos (90{}^\circ -25{}^\circ )}\times \frac{1}{\cos (90{}^\circ -35{}^\circ )} $ $ =\sin 25{}^\circ \times \sin 35{}^\circ \times \frac{1}{\sin 25{}^\circ }\times \frac{1}{\sin 35{}^\circ }=1 $
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