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प्रश्न
If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2
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उत्तर
L.H.S = (m2 + n2) cos2 β
= `((cos^2 alpha)/(cos^2 beta) + (cos^2 alpha)/(sin^2 beta))cos^2 beta`
= `((cos^2 alpha sin^2 beta + cos^2 alpha cos^2 beta)/(cos^2 beta sin^2 beta))cos^2 beta`
= `(cos^2 alpha (sin^2 beta + cos^2 beta) cos^2 beta)/(cos^2 beta sin^2 beta)`
= `(cos^2 alpha (1))/(sin^2 beta)`
= `((cos alpha)/sin beta)^2` = n2
L.H.S = R.H.S
⇒ ∴ (m2 + n2) cos2 β = n2
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= `cosθ/sinθ + square/cosθ`
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∴ L.H.S. = R.H.S.
