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प्रश्न
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
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उत्तर
LHS = cos 1°cos 2°cos 3° ....cos 180°
= cos 1°cos 2°cos 3° ....cos 89° cos 90° .... cos 180°
= cos 1°cos 2°cos 3° ....cos 89° x 0 x cos 91° .... cos 180°
= 0
= RHS
Hence proved.
संबंधित प्रश्न
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 identities.
`(cos A cosec A - sin A sec A)/(cos A + sin A) = cosec A - sec A`
`(sec theta -1 )/( sec theta +1) = ( sin ^2 theta)/( (1+ cos theta )^2)`
If `( sin theta + cos theta ) = sqrt(2) , " prove that " cot theta = ( sqrt(2)+1)`.
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
The value of sin ( \[{45}^° + \theta) - \cos ( {45}^°- \theta)\] is equal to
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
Prove that: `1/(cosec"A" - cot"A") - 1/sin"A" = 1/sin"A" - 1/(cosec"A" + cot"A")`
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
