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प्रश्न
Prove the following:
sec6x – tan6x = 1 + 3sec2x × tan2x
उत्तर
L.H.S. = sec6x – tan6x
L.H.S. = (sec2x)3 – tan6x
L.H.S. = (1 + tan2x)3 – tan6x ...[∵ 1 + tan2θ = sec2θ]
L.H.S. = 1 + 3tan2x + 3(tan2x)2 + (tan2x)3 – tan6x ...[∵ (a + b)3 = a3 + 3a2b + 3ab2 + b3]
L.H.S. = 1 + 3tan2x(1 + tan2x) + tan6x – tan6x
L.H.S. = 1 + 3tan2x.sec2x ...[∵ 1 + tan2θ = sec2θ]
R.H.S. = 1 + 3sec2x.tan2x
L.H.S. = R.H.S.
∴ sec6x – tan6x = 1 + 3sec2x × tan2x
Hence proved.
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संबंधित प्रश्न
If \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tanθ.
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If \[\cot\theta = \frac{40}{9}\], find the values of cosecθ and sinθ.
If tanθ = 1 then, find the value of
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Prove that:
Prove that:
(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.
Prove that:
Prove that: `1/"sec θ − tan θ" = "sec θ + tan θ"`
Choose the correct alternative answer for the following question.
cosec 45° =?
Choose the correct alternative answer for the following question.
Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ
Prove the following.
Prove the following.
Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)
= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`
= `((1 - square)/square) ((square + square)/(square square))`
= `square/square xx 1/(square square)` ......`[(∵ square + square = 1),(∴ square = 1 - square)]`
= `square/(square square)`
= tan θ.sec θ
= R.H.S.
∴ L.H.S. = R.H.S.
∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
