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प्रश्न
Four alternative answers for the following question are given. Choose the correct alternative and write its alphabet:
sin θ × cosec θ = ______
पर्याय
1
0
`1/2`
`sqrt2`
उत्तर
sin θ × cosec θ = 1
Explanation:
sin θ × cosec θ
= `sintheta xx 1/sinθ ... [cosec theta = 1/(sintheta)]`
= 1
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 identities, where the angles involved are acute angles for which the expressions are defined:
`(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`
Prove the following trigonometric identities.
`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`
Prove the following identities:
`(1 - sinA)/(1 + sinA) = (secA - tanA)^2`
Prove that:
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
Prove that:
`1/(sinA - cosA) - 1/(sinA + cosA) = (2cosA)/(2sin^2A - 1)`
`cosec theta (1+costheta)(cosectheta - cot theta )=1`
Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`
Prove the following identity :
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
Prove the following identity :
`(1 + cotA)^2 + (1 - cotA)^2 = 2cosec^2A`
Prove the following identity :
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`
Prove the following identity :
`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`
Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = cosec θ - cot θ`.
Prove that `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1) = (1 + cos "A")/sin "A"`
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
Prove the following identities:
`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.
Prove that: sin4 θ + cos4θ = 1 - 2sin2θ cos2 θ.
Choose the correct alternative:
cot θ . tan θ = ?
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
