Question

If p and q are accute angles such that sinp -cosq= 0,prove that p+q=90°

Answer 1

Given,

P and Q are acute angles.

To prove :-

p + q = 90°

Proof :-

sinP - cosQ = 0

⇒sinP = cosQ

⇒sinP = sin(90°-Q)

(base is same)

⇒P = 90° - Q

⇒P + Q = 90°

Thus proved!

Answer 2

Given: Here we have given p and q are accute angles

To find: Here we have to prove that sinp -cosq= 0,prove that p+q=90°

Solution:

• Here we have given,

$sinP - cosQ = 0\\sinP = cosQ\\sinP = sin(90^o -Q)$

• Multiply both side by $sin^{-1$

$P = 90^o - Q\\P + Q = 90^o$

Hence proved