Advertisements
Advertisements
Question
if `cosec A = sqrt2` find the value of `(2 sin^2 A + 3 cot^2 A)/(4(tan^2 A - cos^2 A))`
Solution
Given `cosec A = sqrt2`
We have to find the value of the expression `(2 sin^2 A + 3 cot^2 A)/(4(tan^2 A - cos^2 A))`
We know that
`cosec A =sqrt2`
`=> sin A = 1/(cosec A) = 1/sqrt2`
`cos A = sqrt(1 - sin^2 A) = sqrt(1 - (1/sqrt2)^2) = 1/sqrt2`
`tan A = sin A/cos A = (1/sqrt2)/(1/sqrt2) = 1`
`cot A = 1/tan A = 1/1 = 1`
Therefore,
`(2 sin^2 A + 3 cot^2 A)/(4(tan^2 A - cos^2 A)) = (2 xx (1/sqrt2)^2 + 3 xx 1^2)/(4(1^2 - (1/sqrt2)^2))`
= 2
Hence, the value of the given expression is 2
APPEARS IN
RELATED QUESTIONS
If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.
Prove the following trigonometric identities.
(cosecA − sinA) (secA − cosA) (tanA + cotA) = 1
if `tan theta = 12/5` find the value of `(1 + sin theta)/(1 -sin theta)`
solve.
cos240° + cos250°
Evaluate.
sin235° + sin255°
Use tables to find cosine of 65° 41’
Use tables to find cosine of 9° 23’ + 15° 54’
If A and B are complementary angles, prove that:
cot B + cos B = sec A cos B (1 + sin B)
Find A, if 0° ≤ A ≤ 90° and 4 sin2 A – 3 = 0
If \[\frac{160}{3}\] \[\tan \theta = \frac{a}{b}, \text{ then } \frac{a \sin \theta + b \cos \theta}{a \sin \theta - b \cos \theta}\]
The value of cos 1° cos 2° cos 3° ..... cos 180° is
In the following Figure. AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot θ.
If sin θ =7/25, where θ is an acute angle, find the value of cos θ.
Prove that:
(sin θ + 1 + cos θ) (sin θ − 1 + cos θ) . sec θ cosec θ = 2
Express the following in term of angles between 0° and 45° :
cosec 68° + cot 72°
Evaluate: `3(sin72°)/(cos18°) - (sec32°)/("cosec"58°)`.
Find the value of the following:
`(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin31^circ) + cos theta/(sin(90^circ - theta))- 8cos^2 60^circ`
Find the value of the following:
`cot theta/(tan(90^circ - theta)) + (cos(90^circ - theta) tantheta sec(90^circ - theta))/(sin(90^circ - theta)cot(90^circ - theta)"cosec"(90^circ - theta))`
Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
Prove the following:
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
