Cosa=cota/coseca=cota/√(1-cot2a)
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Answer:
LHS-
√(1+cosA)/√(1-cosA)
Multiplying with √(1+cosA) on both numerator and denominator, we have,
={√(1+cosA)*√(1+cosA)}/{√(1-cosA)*√(1+cosA)}
=[√{(1+cosA)*(1+cosA)}]/[√{(1-cosA)*(1+cosA)}]
={√(1+cosA)²}/{√(1²-cos²A)}
=(1+cosA)/√(sin²A)
=(1+cosA)/sinA
=(1/sinA)+(cosA/sinA)
=cosecA + cotA
Hence proved…
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