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प्रश्न
Prove the following identities.
tan4 θ + tan2 θ = sec4 θ – sec2 θ
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उत्तर
tan4 θ + tan2 θ = sec4 θ – sec2 θ
L.H.S = tan4 θ + tan2 θ
Taking out tan2 θ as common
= tan2 θ (tan2 θ + 1)
We know that
1 + tan2 θ = sec2 θ
i.e. tan2 θ = sec2 θ - 1
It can be written as
= (sec2 θ – 1) sec2 θ
So we get
= sec4 θ – sec2 θ
= R.H.S
Therefore, it is proved.
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(cos A cosec A - sin A sec A)/(cos A + sin A) = cosec A - sec A`
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = cosec A - cot A`
If 4 cos2 A – 3 = 0, show that: cos 3 A = 4 cos3 A – 3 cos A
If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
If cosA + cos2A = 1, then sin2A + sin4A = 1.
If sin A = `1/2`, then the value of sec A is ______.
(1 – cos2 A) is equal to ______.
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
