हिंदी

From the Figure Find the Value of Sinθ.

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प्रश्न

From the figure find the value of sinθ.

योग
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उत्तर

`sinθ = ("AB")/("AC")`

`sinθ = 3/5`

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2018-2019 (March) Balbharati Model Question Paper Set 1

संबंधित प्रश्न

If acosθ – bsinθ = c, prove that asinθ + bcosθ = `\pm \sqrt{a^{2}+b^{2}-c^{2}`


 

Evaluate

`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`

 

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

`(cosec  θ  – cot θ)^2 = (1-cos theta)/(1 + cos theta)`


Prove the following trigonometric identity.

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


Prove the following trigonometric identities.

`sqrt((1 - cos theta)/(1 + cos theta)) = cosec theta - cot theta`


Prove the following trigonometric identities

`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`


Prove the following identities:

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


If tan A = n tan B and sin A = m sin B, prove that `cos^2A = (m^2 - 1)/(n^2 - 1)`


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


`(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) = 2/((sin^2 theta - cos ^2 theta)) = 2/((2 sin^2 theta -1))`


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


Four alternative answers for the following question are given. Choose the correct alternative and write its alphabet:

sin θ × cosec θ = ______


Simplify 

sin A `[[sinA   -cosA],["cos A"  " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`


Prove the following identity : 

`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`


Find the value of `θ(0^circ < θ < 90^circ)` if : 

`cos 63^circ sec(90^circ - θ) = 1`


Prove that:

tan (55° + x) = cot (35° – x)


Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2. 


`(sin A)/(1 + cos A) + (1 + cos A)/(sin A)` = 2 cosec A


Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`


Prove that:  `1/(sec θ - tan θ) = sec θ + tan θ`.


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