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प्रश्न
cos4 A − sin4 A is equal to ______.
(cos4 A − sin4 A) on simplification, gives
विकल्प
2 cos2 A + 1
2 cos2 A − 1
2 sin2 A − 1
2 sin2 A + 1
उत्तर
cos4 A − sin4 A is equal to 2 cos2 A − 1.
Explanation:
The given expression is cos4 A − sin4 A.
Factorising the given expression, we have
cos4 A − sin4 A = [(cos2 A)2 − (sin2 A)2]
= (cos2 A + sin2 A) × (cos2 A − sin2 A) ...[∵ (a2 − b2) = (a + b)(a − b)]
= cos2 A − sin2 A ...[∵ sin2 A + cos2 A = 1]
= cos2 A − (1 − sin2 A)
= cos2 A − 1 + cos2 A
= 2 cos2 A − 1
संबंधित प्रश्न
if `cos theta = 5/13` where `theta` is an acute angle. Find the value of `sin theta`
Prove the following trigonometric identities
(1 + cot2 A) sin2 A = 1
Prove the following identities:
`(sintheta - 2sin^3theta)/(2cos^3theta - costheta) = tantheta`
Prove the following identities:
`1 - cos^2A/(1 + sinA) = sinA`
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
`(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) = 2/((sin^2 theta - cos ^2 theta)) = 2/((2 sin^2 theta -1))`
Write the value of `(cot^2 theta - 1/(sin^2 theta))`.
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
Write the value of cos1° cos 2°........cos180° .
What is the value of (1 + tan2 θ) (1 − sin θ) (1 + sin θ)?
Write True' or False' and justify your answer the following :
The value of \[\sin \theta\] is \[x + \frac{1}{x}\] where 'x' is a positive real number .
Prove the following identity :
`(cotA + tanB)/(cotB + tanA) = cotAtanB`
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Prove that `sqrt((1 + cos A)/(1 - cos A)) = (tan A + sin A)/(tan A. sin A)`
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
Prove the following identity:
(sin2θ – 1)(tan2θ + 1) + 1 = 0
