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प्रश्न
If a sec θ + b tan θ = m and b sec θ + a tan θ = n, prove that a2 + n2 = b2 + m2.
बेरीज
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उत्तर
a sec θ + b tan θ = m ....(i)
b sec θ + a tan θ = n ....(ii)
Squaring both equations and then subtracting,
a2 sec2 θ + b2 tan2 θ + 2ab sec θ tan θ = m2
b2 sec2 θ + a2 tan2 θ + 2ab sec θ tan θ = n2
− − − −
a2 sec2 θ − b2 sec2 θ + b2 tan2 θ − a2 tan2 θ = m2 − n2
a2 (sec2 θ − tan2 θ) − b2 (sec2 θ − tan2 θ) = m2 − n2
a2 × 1 − b2 × 1 = m2 − n2 ....(sec2 θ − tan2 θ = 1)
a2 + n2 = b2 + m2
Hence Proved.
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