Question

sec² theta .
cos² theta = tan² theta + sin² theta​

Answer 1

Step-by-step explanation:

let theta=x

given sec²x+cos²x=tan²x+sin²x

since sec²x=1/cos²x

1/cos²x+cos²x=(1+cos⁴x)/cos²x=LHS

since tan²x=sin²x/cos²x

sin²x/cos²x +sin²x=(sin²x+sin²xcos²x)/cos²x=RHS

since LHS=RHS

therefore

sin²x(1+cos²x)/cos²x=(1+cos⁴x)/cos²x

sin²x(1+cos²x)=(1+cos⁴x)

since 1+cos²x=sin²x

therefore

sin⁴x=1+cos⁴x

sin⁴x-cos⁴x=

(sin²x+cos⁴x)(sin²x-cos²x)=1

sin²x-cos²x=1

which is only possible if x=90° or x=0°

therefore x=(2n+1)π/2 or x=nπ