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महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

If 1 – cos^2θ = 1/4, then θ = ?

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प्रश्न

If `1 - cos^2θ = 1/4`, then θ = ?

बेरीज
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उत्तर

`1 - cos^2θ = 1/4`   ...[Given]

∴ `sin^2θ = 1/4`   ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 - cos^2θ = sin^2θ)]`

∴ `sin θ = 1/2`   ...[Taking square root of both sides]

∴ θ = 30°   ...`[∵ sin 30^circ = 1/2]`

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Trigonometry - Q.1 (B)

संबंधित प्रश्‍न

Prove the following trigonometric identities.

`(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`


Prove the following trigonometric identities.

`(tan^2 A)/(1 + tan^2 A) + (cot^2 A)/(1 + cot^2 A) = 1`


Prove the following identities:

`(1 - sinA)/(1 + sinA) = (secA - tanA)^2`


Prove the following identities:

`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (1 + cosA)/sinA`


Prove that:

`sqrt(sec^2A + cosec^2A) = tanA + cotA`


`tan theta /((1 - cot theta )) + cot theta /((1 - tan theta)) = (1+ sec theta cosec  theta)`


Write the value of `(1 - cos^2 theta ) cosec^2 theta`.


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Prove the following identity :

`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`


Prove the following identity : 

`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`


If sinA + cosA = m and secA + cosecA = n , prove that n(m2 - 1) = 2m


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.


If A = 60°, B = 30° verify that tan( A - B) = `(tan A - tan B)/(1 + tan A. tan B)`.


If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.


cos θ . sec θ = ?


Prove that cot2θ – tan2θ = cosec2θ – sec2θ.


Which of the following is true for all values of θ (0° ≤ θ ≤ 90°)?


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