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
संबंधित प्रश्न
Prove the following trigonometric identities.
`((1 + cot^2 theta) tan theta)/sec^2 theta = cot theta`
Evaluate.
cos225° + cos265° - tan245°
For triangle ABC, show that : `tan (B + C)/2 = cot A/2`
Prove that:
`(sinthetasin(90^circ - theta))/cot(90^circ - theta) = 1 - sin^2theta`
What is the maximum value of \[\frac{1}{\sec \theta}\]
If A + B = 90°, then \[\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}\]
Prove that:
\[\left( \frac{\sin49^\circ}{\cos41^\circ} \right)^2 + \left( \frac{\cos41^\circ}{\sin49^\circ} \right)^2 = 2\]
Express the following in term of angles between 0° and 45° :
cosec 68° + cot 72°
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
cos(90° - A) · sec 77° = 1
Find the value of the following:
tan 15° tan 30° tan 45° tan 60° tan 75°
