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प्रश्न
Evaluate the following: `(3sin^2 40°)/(4cos^2 50°) - ("cosec"^2 28°)/(4sec^2 62°) + (cos10° cos25° cos45° "cosec"80°)/(2sin15° sin25° sin45° sin65° sec75°)`
बेरीज
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उत्तर
`(3sin^2 40°)/(4cos^2 50°) - ("cosec"^2 28°)/(4sec^2 62°) + (cos10° cos25° cos45° "cosec"80°)/(2sin15° sin25° sin45° sin65° sec75°)`
= `(3sin^2 (90° - 50°))/(4cos^2 50°) - ("cosec"^2 (90° - 62°))/(4sec^2 62°) + (cos(90° - 80°) cos25° xx 1/sqrt(2) xx 1/(sin80°))/(2sin(90° - 75°) xx 1/sqrt(2) xx sin(90° - 25°) xx 1/(cos75°))`
= `(3cos^2 50°)/(4cos^2 50°) - (sec^2 62°)/(4sec^2 62°) + (sin80° xx cos25° xx 1/(cos75°))/(2cos75° xx cos25° xx 1/(cos75°))`
= `(3)/(4) - (1)/(4) + (1)/(2)`
= `(1)/(2) + (1)/(2)`
= 1.
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Trigonometric Equation Problem and Solution
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