prove that (1+tan^A/1+cot^A)=(1-tan A /1-cot A)^=tan^A
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Answered by
16
hello there!!!!!
we have,
LHS= (1+tan²A/1+cot²A)
=> sec²A/cosec²A [SINCE sec²A=1+tan²A, cosec²A=1+cot²A]
=> 1/cos²A/1/sin²A [since 1/cosA=secA ,1/sinA=cosecA]
=> sin²A/cos²A=tan²A= RHS
middle term= (1-tanA/1-cotA)²
=> [ (1-sinA/cosA)/(1-cosA/sinA)]² [since tanA=sinA/cosA ,cotA=cosA/sinA]
=>(cosA-sinA/cosA)/(sinA-cosA/sinA)]² [taking LCM]
=> [sinA(cosA-sinA)/cosA(sinA-cosA)]²
=> [ -sinA(sinA-cosA)/cosA(sinA-cosA)]²
=> (-sinA/cosA)²= (-tanA)² = tan²A =RHS
we have,
LHS= (1+tan²A/1+cot²A)
=> sec²A/cosec²A [SINCE sec²A=1+tan²A, cosec²A=1+cot²A]
=> 1/cos²A/1/sin²A [since 1/cosA=secA ,1/sinA=cosecA]
=> sin²A/cos²A=tan²A= RHS
middle term= (1-tanA/1-cotA)²
=> [ (1-sinA/cosA)/(1-cosA/sinA)]² [since tanA=sinA/cosA ,cotA=cosA/sinA]
=>(cosA-sinA/cosA)/(sinA-cosA/sinA)]² [taking LCM]
=> [sinA(cosA-sinA)/cosA(sinA-cosA)]²
=> [ -sinA(sinA-cosA)/cosA(sinA-cosA)]²
=> (-sinA/cosA)²= (-tanA)² = tan²A =RHS
Anonymous:
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here is no any doubt ...thanks
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Answered by
8
Answer is Option D tan²A
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