Advertisements
Advertisements
प्रश्न
If A and B are complementary angles, prove that:
cot A cot B – sin A cos B – cos A sin B = 0
Advertisements
उत्तर
Since, A and B are complementary angles, A + B = 90°
cot A cot B – sin A cos B – cos A sin B
= cot A cot (90° – A) – sin A cos (90° – A) – cos A sin (90° – A)
= cot A tan A – sin A sin A – cos A cos A
= 1 – (sin2 A + cos2 A)
= 1 – 1
= 0
APPEARS IN
संबंधित प्रश्न
If A, B, C are the interior angles of a triangle ABC, prove that `\tan \frac{B+C}{2}=\cot \frac{A}{2}`
if `cot theta = 1/sqrt3` find the value of `(1 - cos^2 theta)/(2 - sin^2 theta)`
Evaluate:
`3 sin72^circ/(cos18^circ) - sec32^circ/(cosec58^circ)`
A triangle ABC is right angles at B; find the value of`(secA.cosecC - tanA.cotC)/sinB`
Prove that:
`(cos(90^circ - theta)costheta)/cottheta = 1 - cos^2theta`
Use tables to find the acute angle θ, if the value of sin θ is 0.6525
\[\frac{2 \tan 30° }{1 + \tan^2 30°}\] is equal to
\[\frac{2 \tan 30°}{1 - \tan^2 30°}\] is equal to ______.
Evaluate:
3 cos 80° cosec 10°+ 2 sin 59° sec 31°
In ∆ABC, `sqrt(2)`AC = BC, sin A = 1, sin2A + sin2B + sin2C = 2, then ∠A = ?, ∠B = ?, ∠C = ?
