solution required as early as possible...

Asked by ABHILASHA | 5th Mar, 2019, 11:20: PM

Expert Answer:

begin mathsize 16px style fraction numerator sin cubed straight theta space plus space cos cubed straight theta over denominator sinθ space plus space cosθ end fraction
Since space left parenthesis straight a cubed space plus space straight b cubed right parenthesis space equals space left parenthesis space straight a space plus space straight b right parenthesis space left parenthesis straight a squared space plus space straight b squared space minus space ab right parenthesis straight p
fraction numerator begin display style sin cubed straight theta space plus space cos cubed straight theta end style over denominator begin display style sinθ space plus space cosθ end style end fraction equals space space fraction numerator open parentheses sinθ space plus space cosθ close parentheses begin display style open parentheses sin squared straight theta space plus space cos squared straight theta space minus space sinθcosθ close parentheses end style over denominator open parentheses sinθ space plus space cosθ close parentheses end fraction equals 1 space minus space space sinθcosθ space end style

Answered by Yasmeen Khan | 6th Mar, 2019, 10:33: AM