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प्रश्न
Prove the following trigonometric identities.
`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`
उत्तर
We have to prove `tan theta - cot theta = (2 sin^2 theta - `1)/(sin theta cos theta)`
We know that. `sin^2 theta + cos^2 theta - 1`
So,
`tan theta - cot theta = sin theta/cos theta - cos theta/sin theta`
`= (sin^2 theta - cos^2 theta)/(sin theta cos theta)`
`= (sin^2 theta - (1 - sin^2 theta))/(sin theta cos theta)`
`= (sin^2 theta - (1 - sin^2 theta))/(sin theta cos theta)`
`= (2 sin^2 theta - 1)/(sin theta cos theta)`
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संबंधित प्रश्न
Prove the following identities:
`(i) cos4^4 A – cos^2 A = sin^4 A – sin^2 A`
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Prove that:
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Prove the following identities:
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If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = 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^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Show that tan4θ + tan2θ = sec4θ – sec2θ.
