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Prove the Following.Sec2θ + Cosec2θ = Sec2θ × Cosec2θ - Geometry Mathematics 2

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प्रश्न

Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ 

बेरीज
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उत्तर

\[\sec^2 \theta + {cosec}^2 \theta\]
\[ = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}\]
\[ = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta}\]
\[ = \frac{1}{\cos^2 \theta \sin^2 \theta}\]
\[ = \frac{1}{\cos^2 \theta} \times \frac{1}{\sin^2 \theta}\]
\[ = \sec^2 \theta  \text{ cosec }^2 \theta\]
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पाठ 6: Trigonometry - Problem Set 6 [पृष्ठ १३८]

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बालभारती Mathematics 2 [English] Standard 10 Maharashtra State Board
पाठ 6 Trigonometry
Problem Set 6 | Q 5.03 | पृष्ठ १३८

संबंधित प्रश्‍न

If \[\tan \theta = \frac{3}{4}\], find the values of sec​θ and cos​θ


Prove that:

cos2θ (1 + tan2θ)


Prove that:

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

Prove that:

(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.


Prove that:

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

Prove that:

\[\sec^4 A\left( 1 - \sin^4 A \right) - 2 \tan^2 A = 1\]

Choose the correct alternative answer for the following question.
sin \[\theta\] cosec \[\theta\]= ?


Choose the correct alternative answer for the following question.
cosec 45° =?


Choose the correct alternative answer for the following question.

1 + tan2 \[\theta\]  = ?


Choose the correct alternative answer for the following question.

When we see at a higher level, from the horizontal line, angle formed is ........
 

Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ


Prove the following:

sec6x – tan6x = 1 + 3sec2x × tan2x


Prove the following.

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


In ΔPQR, ∠P = 30°, ∠Q = 60°, ∠R = 90° and PQ = 12 cm, then find PR and QR.


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


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