question b how to solve it
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We are given, LHS = √sec² A + cosec² A
= √1+tan² A + 1+ cot² A [∵sec²A = 1+tan²A and cosec²A = 1+cot²A]
= √tan²A + cot²A + 2(1)
= √tan²A + cot²A + 2(tanA*cotA) [∵tanA*cotA =1]
= √(tanA+cot)² [a²+b²+2ab = (a+b)²]
= tanA + cot A = RHS
= √1+tan² A + 1+ cot² A [∵sec²A = 1+tan²A and cosec²A = 1+cot²A]
= √tan²A + cot²A + 2(1)
= √tan²A + cot²A + 2(tanA*cotA) [∵tanA*cotA =1]
= √(tanA+cot)² [a²+b²+2ab = (a+b)²]
= tanA + cot A = RHS
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this is very simple simplify both LHS and RHS
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mahi10august:
you have got two ways to answer this ques
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