Question

. Show that sin (90 - 0) = cos O​

Answer 1

Answer:

Can put other values too.

this is only 1 method.... there are still many you can use

Answer 2

To prove

Sin(90-Θ)=cosΘ

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

From above figure

S stands for hypotenuse

h stands for height

and l is the base or length of BC

★SinΘ=perpendicular/hypotenuse (p/h)

★CosΘ=Base/height (b/h)

★TanΘ=Perpendicular/base

From above figure:-

SinΘ=h/s

90- a must be equal to h (adjacent side)

Cos(90-Θ)=h/s

We don't take l/s because 90-Θ is an angle and l will be the perpendicular as it lies just front of it.

=>we know that the side in front of an angle will be considered as perpendicular.

So,cosΘ=b/h and base would be h

If sinΘ = h/s and cos(90-Θ)=h/s

Then sinΘ must be equal to cos(90-Θ)

Therefore,sinΘ=cos(90-Θ)(proved)

Alternative Method

Sin (a-b)=sinacosb-sinbcosa

Replace with sin(90-Θ)

Sin(90-Θ)= sin90cosΘ-sinacos90

We know Sin90°= 1 and cos90°= 0

Sin(90-Θ)= 1(cosΘ)-sina(0)

Sin(90-Θ)=cosΘ (proved)