Advertisements
Advertisements
Question
Prove the following trigonometric identities.
(cosecA − sinA) (secA − cosA) (tanA + cotA) = 1
Solution
We have to prove (cosecA − sinA) (secA − cosA) (tanA + cotA) = 1
We know that `sin^2 A + cos^2 A = 1`
So
(cosec A - sin A)(sec A - cos A)(tan A + cot A)
`= (1/sin A - sin A)(1/cos A - cos A)(sin A/cos A + cos A/sin A)`
`= ((1 - sin^2 A)/sin A) ((1 - cos^2 A)/cos A) (sin^2 A + cos^2 A)/(sin A cos A)`
`= ((cos^2 A)/sin A) ((sin^2 A)/cos A) (1/(sin A cos A))`
`(= sin^2 A cos^2 A)/(sin^2 A cos^2 A)`
= 1
APPEARS IN
RELATED QUESTIONS
Evaluate `(tan 26^@)/(cot 64^@)`
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Prove the following trigonometric identities.
(cosecθ + sinθ) (cosecθ − sinθ) = cot2 θ + cos2θ
Evaluate.
`cos^2 26^@+cos65^@sin26^@+tan36^@/cot54^@`
Express the following in terms of angles between 0° and 45°:
cos74° + sec67°
For triangle ABC, show that : `sin (A + B)/2 = cos C/2`
Prove that:
sin (28° + A) = cos (62° – A)
If A and B are complementary angles, prove that:
cot B + cos B = sec A cos B (1 + sin B)
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that cos 3 A = 4 cos3 A – 3 cos A
If 0° < A < 90°; find A, if `(cos A )/(1 - sin A) + (cos A)/(1 + sin A) = 4`
If 0° < A < 90°; find A, if `sinA/(secA - 1) + sinA/(secA + 1) = 2`
∠ACD is an exterior angle of Δ ABC. If ∠B = 40o, ∠A = 70o find ∠ACD.
Sin 2A = 2 sin A is true when A =
If A, B and C are interior angles of a triangle ABC, then \[\sin \left( \frac{B + C}{2} \right) =\]
Prove that:
\[\frac{sin\theta \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]
Evaluate: cos2 25° - sin2 65° - tan2 45°
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
cos(90° - A) · sec 77° = 1