Question

if sinA=cosB then (A+B) = 90°​

Answer 1

Given :

• sin A = cos B

To Prove :

• A + B = 90°

Solution :

We are given,

sin A = cos B

Now, we know that,

$\Large \underline{\boxed{\bf{ cos \theta = sin (90^{\circ} - \theta) }}}$

$\sf : \implies sin A = sin (90^{\circ} - B)$

$\sf : \implies A = 90^{\circ} - B$

$\sf : \implies A + B = 90^{\circ}$

$\Large \underline{\boxed{\bf{A + B = 90^{\circ}}}}$

Hence, Proved.

Answer 2

Answer:

Given :

sin A = cos B

To Prove :

A + B = 90°

Solution :

We are given,

sin A = cos B

Now, we know that,

cos \theta = sin (90 - theta)

cosθ=sin(90−θ)

sin A = sin (90- B):⟹sinA=sin(90∘

−B)

A = 90- B:⟹A=90∘ −B

A + B = 90⟹A+B=90

A + B = 90

Hence, Proved.