Question

Prove that: sin A cos (90° - A) + sin (90°- A) cos A = 1​

Answer 1

Answer:

cos(90−A)=sin(A);

sin(90−A)=cos(A);

tan(90−A)=cot(A)

cot(A)=

sin(A)

cos(A)

LHS

tan(90−A)

cos(90−A)sin(90−A)

=

cot(A)

sin(A)cos(A)

cot(A)

sin(A)cos(A)

=

cos(A)

sin(A)cos(A)sin(A)

cos(A)

sin(A)cos(A)sin(A)

=sin(A)sin(A)=sin

2

(A)=LHS

RHS=sin

2

(A)

LHS=RHS

Hence Proved

Answer 2

Proof:-

Sin(A)=cos(90-A)

cos(A)=sin(90-A)

THEREFORE,sin(A)cos(90-A)=sin(A)^2

AND, sin(90-A)cos(A)=sin(90-A)^2

Therefore, sin(A)^2+sin(90-A)^2

=sin(A)^2+cos(A)^2

Therefore, by Pythagorus theorem(i cant draw the diagram here but try to understand)

sin(A)^2+cos(A)^2=1.

There might be a formula in your textbook for this one:-

sin(theta)^2+cos(theta)^2=1.

HENCE PROOVED