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प्रश्न
Prove the following identities:
`tan A - cot A = (1 - 2cos^2A)/(sin A cos A)`
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उत्तर
L.H.S. = tan A – cot A
= `(sin A)/(cos A) - (cos A)/(sin A)`
= `(sin^2A - cos^2A)/(sin A cos A)`
= `(1 - cos^2A - cos^2A)/(sin A cos A)` ...(∵ sin2A = 1 – cos2A)
= `(1 - 2cos^2A)/(sin A cos A)`
= R.H.S.
संबंधित प्रश्न
if `a cos^3 theta + 3a cos theta sin^2 theta = m, a sin^3 theta + 3 a cos^2 theta sin theta = n`Prove that `(m + n)^(2/3) + (m - n)^(2/3)`
Prove that:
`tanA/(1 - cotA) + cotA/(1 - tanA) = secA "cosec" A + 1`
If `cosA/cosB = m` and `cosA/sinB = n`, show that : (m2 + n2) cos2 B = n2.
Prove the following identities:
`1/(cosA + sinA) + 1/(cosA - sinA) = (2cosA)/(2cos^2A - 1)`
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
Write the value of cos1° cos 2°........cos180° .
If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1
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
The value of 2sinθ can be `a + 1/a`, where a is a positive number, and a ≠ 1.
