Under root sec theta minus 1 …
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Sia ? 6 years ago
LHS =√secθ−1secθ+1+√secθ+1secθ−1
Rationalise the denominator and we get,
=√(secθ−1)2sec2θ−1+√(secθ+1)2sec2θ−1
= (secθ−1)+(secθ+1)√(secθ+1)(secθ−1)
= 2secθ√sec2θ−1=2secθ√tan2θ=2secθtanθ
= 2×1cosθ×cosθsinθ
= 2×1sinθ
= 2 cosecθ
= RHS
Hence Proved
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