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cot inverse (√1+sinx +√1- sinx/√1+sinx - …

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cot inverse (√1+sinx +√1- sinx/√1+sinx - √1-sinx)

Posted by Manjinder Waheguru Singh 6 years, 5 months ago

CBSE > Class 12 > Mathematics

1 answers

Suman Bharti 6 years, 5 months ago

cot inverse [√(cos^x/2+sin^x/2+2sinx/2.cosx/2)+√(cos^x/2+sin^x/2-2sinx/2.cosx/2)/√(cos^x/2+sin^x/2+2sinx/2.cosx/2)-√(cos^x/2+sin^x/2-2sinx/2.cosx/2)] =cot inverse [√(cosx/2+sinx/2)^2+√(cosx/2-sinx/2)^2/√(cosx/2+sinx/2)^2-√(cosx/2-sinx/2)^2] =cot inverse [(cosx/2+sinx/2+cosx/2-sinx/2)/(cosx/2+sinx/2-cosx/2+sinx/2)] =cot inverse [(2cosx/2)/(2sinx/2)] =cot inverse (cotx/2) =x/2

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