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प्रश्न
Prove that:
cos2θ (1 + tan2θ)
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उत्तर
L.H.S. = cos2θ (1 + tan2θ)
= cos2θ × sec2θ ...[∵ 1 + tan2 θ = sec2 θ]
= \[\cos^{2}\theta\times\frac{1}{\cos^{2}\theta}\]
= 1
= R.H.S.
∴ cos2θ (1 + tan2θ) = 1
संबंधित प्रश्न
If \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tanθ.
If tanθ = 1 then, find the value of
`(sinθ + cosθ)/(secθ + cosecθ)`
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If \[\tan\theta + \frac{1}{\tan\theta} = 2\], then show that \[\tan^2 \theta + \frac{1}{\tan^2 \theta} = 2\]
Prove the following.
secθ (1 – sinθ) (secθ + tanθ) = 1
Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ
Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ
Prove the following:
sec6x – tan6x = 1 + 3sec2x × tan2x
Prove the following.
\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]
Prove the following.
Choose the correct alternative:
sinθ × cosecθ =?
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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
