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Question
`(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin 31^circ) - 8 sin^2 30^circ` is equal to ______.
Options
1
– 1
0
2
MCQ
Fill in the Blanks
Solution
`(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin 31^circ) - 8 sin^2 30^circ` is equal to 0.
Explanation:
We have, `(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin 31^circ) - 8 sin^2 30^circ`
= `(sin 20^circ)/(sin 20^circ) + (sin 31^circ)/(sin 31^circ) - 8(1/2)^2` ...[∵ cos θ = sin(90° – θ), sin 30° = 1/2]
= `1 + 1 - 8 xx 1/4`
= 2 – 2
= 0
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Trigonometric Functions of Sum and Difference of Angles
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