Question

(SinA+cosecA)^2+(cosA+secA)^2-(tan^2A+cot^2A)=7

Answer 1

Answer:

Step-by-step explanation:

l.h.s.

(sina+coseca)^2+(cosa+seca)^2

= (sin2a + cosec2a + 2sina.coseca)+(cos2a+sec2a+2seca.cosa).........(a+b)^2 formula

= (sin2a + cosec2a+2sina.1/sina)+(cos2a+sec2a+2cosa.1/cosa)

= (sin2a + cosec2a+2)+(cos2a+sec2a+2)

= (sin2a+cos2a)+4+ cosec2a+sec2a)

= 1+ 4 +( 1 + cot2a) +(1+ tan2a).........cosec2a=( 1 + cot2a) & sec2a=+(1+ tan2a)

= 7+ cot2a+ tan2a

l.h.s.

(sina+coseca)^2+(cosa+seca)^2-tan2a-cot2a= 7

proved......

Answer 2

$( {sina + coseca})^{2} + ( {cosa + seca})^{2} - ( {tan}^{2} a + {cot}^{2} a) \\ \\ = {sin}^{2} a + {cosec}^{2} a + 2 + {cos}^{2} a + {sec}^{2} a + 2 - {tan}^{2} a - {cot}^{2} a \\ \\ = 5 + 1 + {cot}^{2} a + 1 + {tan}^{2} a - {tan}^{2} a - {cot}^{2} a \\ \\ = 7$