Question

Prove that :– (sec a + cos a )(sec a - Cos a) = (tan square a + sin square a)​

Answer 1

Step-by-step explanation:

taking l.h.s

=(sec a+cos a)(sec a- cos a)

since,(a+b)(a-b)=a square-b square

=sec square a - cos square a

since, sec a = 1/cos a

therefore, sec square a=1/cos square a

=1/cos square a - cos square a

taking lcm

=[1 - (cos square a)square]/cos square a

since,1 can be written as 1 square

=[1 square -(cos square a)square]/cos square a

since, a square - b square = (a+b)(a-b)

=[(1-cos square a)(1+cos square a)]/cos square a

since,1- cos square a =sin square a

=sin square a(1+cos square a)/cos square a

=sin square a+ sin square a cos square a/cos square a

=sin square a/cos square a + sin square a cos square a/ cos square a

since,sin square a / cos square a=tan square a

tan square a+ sin square a

LHS=RHS

hence proved