prove secA-tanA \secA+tanA the whole root -cosecA-cotA\cosecA+cotA the whole root =secA-1 cosecA-1
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a bdbdjhddbsnsns is 3737:is the right answer
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We have,
√SecA-tan A/SecA+tana=√SecA-tanA/√SecA+tanA×√SecA-tanA/SecA-tanA
=(SecA-tanA) /(Sec^2A-tan^2A) =(SecA-tanA) ------- equation 1 And
√cosec A- cotA /cosec A+cot A=√cosec A - cot A /√cosec A +cot A×√√cosec A-cotA/√cosec A-cotA
=(cosec A- cot A) /√(cosec^2-cot^2A) ----- equation 2
[cosec ^2A - cot ^2=1]
From equation 1 and 2 we get.
=(1-sinA ) (1-cos A) /cos A sin A=(1-cosA) /cosA . (1-sinA) /sinA
=(1/cosA-1)(1/sinA-1)
=(secA-1) (cosec A -1) =RHS.
Therefore, LHS=RHS.
Step-by-step explanation:
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