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प्रश्न
Prove the following:
`(sin3x)/cosx + (cos3x)/sinx` = 2 cot 2x
बेरीज
उत्तर
L.H.S. = `(sin3x)/cosx + (cos3x)/sinx`
= `(sin3x*sinx + cos3x*cosx)/(sinx*cosx)`
= `(cos(3x - x))/(sinx*cosx)`
= `(2cos2x)/(2sinx*cosx)`
= `(2cos2x)/(sin2x)`
= 2 cot 2x
= R.H.S.
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Trigonometric Functions of Double Angles
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometry - 2 - Exercise 3.3 [पृष्ठ ४८]
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