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प्रश्न
`cosec theta (1+costheta)(cosectheta - cot theta )=1`
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उत्तर
LHS = `cosec theta (1+ cos theta )( cosec theta - cot theta)`
=` (cosec theta + cosec theta xx cos theta)(cosec theta - cot theta)`
=` (cosec theta + 1/(sin theta) xx cos theta ) ( cosec theta - cot theta )`
=` ( cosec theta + cot theta )( cosec theta - cot theta)`
=` cosec^2 theta - cot^2 theta (∵ cosec^2 theta - cot^2 theta=1)`
= 1
= RHS
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + cos A)/sin^2 A = 1/(1 - cos A)`
Prove the following trigonometric identities.
(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
Prove the following trigonometric identities.
if cos A + cos2 A = 1, prove that sin2 A + sin4 A = 1
If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
Prove the following identities:
(sec A – cos A) (sec A + cos A) = sin2 A + tan2 A
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
`sin^2 theta + cos^4 theta = cos^2 theta + sin^4 theta`
Write the value of `sin theta cos ( 90° - theta )+ cos theta sin ( 90° - theta )`.
If tan A =` 5/12` , find the value of (sin A+ cos A) sec A.
Given `cos38^circ sec(90^circ - 2A) = 1` , Find the value of <A
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
Without using trigonometric identity , show that :
`cos^2 25^circ + cos^2 65^circ = 1`
Prove that: 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = 0.
There are two poles, one each on either bank of a river just opposite to each other. One pole is 60 m high. From the top of this pole, the angle of depression of the top and foot of the other pole are 30° and 60° respectively. Find the width of the river and height of the other pole.
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
If sin θ (1 + sin2 θ) = cos2 θ, then prove that cos6 θ – 4 cos4 θ + 8 cos2 θ = 4
If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ
sin2θ + sin2(90 – θ) = ?
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 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")`
