Question

Please answer it as soon as possible. Q.no. 8

Answer 1

Answer:

obviously 1,please mark as brainliest

Answer 2

$\boxed{\boxed{\huge{\bf{Solution}}}}$

This question is simply related to trigonometric identities.

We have to determine

sin²∅ + cos²∅

We clearly can see that the given ∆DBC is a right triangle

So,

CD² = BC² + BD²

Now we will apply trigonometric ratios here.

As we know

sin∅

= perpendicular/hypotenuse

= BD/CD

And

cos∅

= base/hypotenuse

= BC/CD

Now,

sin²∅ + cos²∅

= (BD/CD)² + (BC/CD)²

= BD²/CD² + BC²/CD²

Taking LCM

= (BD² + BC²)/CD²

(Now by pythogaurous theorem in ∆DBC,

CD² = BC² + BD²)

= CD²/CD²

= 1

So, sin²∅ + cos²∅ = 1