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प्रश्न
Prove the following:
`cos (pi/2 - x) cos(pi/2 - y) - sin(pi/2 - x) sin(pi/2 - y)` = – cos (x + y)
योग
उत्तर
L.H.S. = `cos (pi/2 - x) cos(pi/2 - y) - sin(pi/2 - x) sin(pi/2 - y)`
= sin x sin y – cos x cos y ...`[(because cos(pi/2 - theta) = sintheta),(sin(pi/2 - theta) = costheta)]`
= – (cos x cos y – sin x sin y)
= – cos (x + y)
= R.H.S.
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Trigonometric Functions of Multiple Angles
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