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Prove that: cos2θ (1 + tan2θ) - Geometry Mathematics 2

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प्रश्न

Prove that:

cos2θ (1 + tan2θ)

बेरीज
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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

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पाठ 6: Trigonometry - Practice Set 6.1 [पृष्ठ १३१]

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बालभारती Mathematics 2 [English] Standard 10 Maharashtra State Board
पाठ 6 Trigonometry
Practice Set 6.1 | Q 6.02 | पृष्ठ १३१

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If 5 secθ – 12 cosecθ = 0, find the values of secθ, cosθ, and sinθ.


If tanθ = 1 then, find the value of

`(sinθ + cosθ)/(secθ + cosecθ)`


Prove that:

\[\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} = \sec\theta - \tan\theta\]

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\[\sec\theta + \tan\theta = \frac{\cos\theta}{1 - \sin\theta}\]

Choose the correct alternative answer for the following question.
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Prove the following.

\[\frac{1}{1 - \sin\theta} + \frac{1}{1 + \sin\theta} = 2 \sec^2 \theta\]

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Prove the following.

\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]


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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))`

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 = `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


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