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प्रश्न
Prove the following identity :
`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`
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उत्तर
LHS = `2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1`
= `2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1`
= `2[(sin^2θ)^3 + (cos^2θ)^3] - 3(sin^4θ + cos^4θ) + 1`
= `2[(sin^2θ + cos^2θ){(sin^2θ)^2 + (cos^2θ)^2 - sin^2θcos^2θ}] - 3(sin^4θ + cos^4θ) + 1`
= `2{(sin^2θ)^2 + (cos^2θ)^2 - sin^2θcos^2θ} - 3(sin^4θ + cos^4θ) + 1`
= `2sin^4θ + 2cos^4θ - 2sin^2θcos^2θ - 3sin^4θ - 3cos^4θ + 1`
= `-sin^4θ - cos^4θ - 2sin^2θcos^2θ + 1`
= `-(sin^4θ + cos^4θ + 2sin^2θcos^2θ) + 1`
= `-(sin^2θ + cos^2θ)^2 + 1 = -1 + 1 = 0`
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संबंधित प्रश्न
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If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
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Prove that sec2 (90° - θ) + tan2 (90° - θ) = 1 + 2 cot2 θ.
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.
Prove that `(cot A)/(1 - cot A) + (tan A)/(1 - tan A) = -1`.
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
