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प्रश्न
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
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उत्तर
LHS = `(cosecθ)/(tanθ + cotθ)`
= `(1/sinθ)/(sinθ/cosθ + cosθ/sinθ)`
= `(1/sinθ)/((sin^2θ + cos^2θ)/(cosθsinθ))` = `(1/sinθ)/(1/(cosθsinθ)`
= `1/sinθ xx (cosθsinθ)/1 = cosθ`
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संबंधित प्रश्न
If (secA + tanA)(secB + tanB)(secC + tanC) = (secA – tanA)(secB – tanB)(secC – tanC) prove that each of the side is equal to ±1. We have,
(secA + tanA) (1 − sinA) = ______.
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
Prove the following trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
If x = h + a cos θ, y = k + b sin θ.
Prove that `((x - h)/a)^2 + ((y - k)/b)^2 = 1`.
`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.
Activity:
`5/(sin^2θ) - 5cot^2θ`
= `square (1/(sin^2θ) - cot^2θ)`
= `5(square - cot^2θ) ...[1/(sin^2θ) = square]`
= 5(1)
= `square`
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
