Advertisements
Advertisements
Question
If \[\sin \theta = \frac{1}{3}\] then find the value of 2cot2 θ + 2.
Advertisements
Solution
Given:
`sin θ=1/3`
⇒ `1/ sinθ=3`
⇒` cosec θ=3`
We know that,
`cosec^2θ-cot ^2θ=1`
⇒`(3)^2-cot^2θ=1`
⇒ `cot ^2 θ=9-1`
Therefore,
`2 cot ^2 θ+2=2xx8+2`
=`16+2`
= `18`
APPEARS IN
RELATED QUESTIONS
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Prove the following identities:
`cosecA - cotA = sinA/(1 + cosA)`
Prove that:
cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
`tan theta/(1+ tan^2 theta)^2 + cottheta/(1+ cot^2 theta)^2 = sin theta cos theta`
`(cos ec^theta + cot theta )/( cos ec theta - cot theta ) = (cosec theta + cot theta )^2 = 1+2 cot^2 theta + 2cosec theta cot theta`
Show that none of the following is an identity:
(i) `cos^2theta + cos theta =1`
If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =
Prove the following identity :
cosecθ(1 + cosθ)(cosecθ - cotθ) = 1
Prove the following identity :
`sec^2A.cosec^2A = tan^2A + cot^2A + 2`
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
If x = asecθ + btanθ and y = atanθ + bsecθ , prove that `x^2 - y^2 = a^2 - b^2`
Without using trigonometric table , evaluate :
`cosec49°cos41° + (tan31°)/(cot59°)`
If sec θ + tan θ = m, show that `(m^2 - 1)/(m^2 + 1) = sin theta`
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
Prove that `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2 = 1`
If A + B = 90°, show that `(sin B + cos A)/sin A = 2tan B + tan A.`
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` ...`[sin^2"A" + square = 1]`
= `square` – cos2A ...[sin2A = 1 – cos2A]
= `square`
= R.H.S.
Prove that `(1 + sin B)/(cos B) + (cos B)/(1 + sin B) = 2 sec B`.
