Advertisements
Advertisements
प्रश्न
Prove the following
sin θ sin (90° − θ) − cos θ cos (90° − θ) = 0
उत्तर
We have to prove sin θ sin (90° − θ) − cos θ cos (90° − θ) = 0
Left-hand side
`= sin theta.sin(90^@ - theta) - cos theta.cos(90^@ - theta)`
`= sin theta.cos theta - cos thete.sin theta`
`= sin theta (cos theta -cos theta)`
= 0
=Right hand side
Proved
APPEARS IN
संबंधित प्रश्न
Express each one of the following in terms of trigonometric ratios of angles lying between
0° and 45°
Sin 59° + cos 56°
Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°
tan 65° + cot 49°
If A, B, C are the interior angles of a triangle ABC, prove that
`tan ((C+A)/2) = cot B/2`
Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.
sec78° + cosec56°
If cos 20 = sin 4 θ ,where 2 θ and 4 θ are acute angles, then find the value of θ
Without using tables, evaluate the following: sin60° sin30°+ cos30° cos60°
Verify the following equalities:
cos 90° = 1 – 2sin2 45° = 2cos2 45° – 1
If sin 30° = x and cos 60° = y, then x2 + y2 is
The value of 5 sin2 90° – 2 cos2 0° is ______.
Evaluate: `(5cos^2 60° + 4sec^2 30° - tan^2 45°)/(sin^2 30° + sin^2 60°)`
