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महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

Prove the Following.(Secθ + Tanθ) (1 – Sinθ) = Cosθ

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प्रश्न

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

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उत्तर

\[\left( \sec\theta + \tan\theta \right)\left( 1 - \sin\theta \right)\]

\[ = \left( \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} \right)\left( 1 - \sin\theta \right)\]

\[ = \left( \frac{1 + \sin\theta}{\cos\theta} \right)\left( 1 - \sin\theta \right)\]

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

\[ = \frac{\cos^2 \theta}{\cos\theta} \left( \sin^2 \theta + \cos^2 \theta = 1 \right)\]

\[ = \cos\theta\]

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पाठ 6: Trigonometry - Problem Set 6 [पृष्ठ १३८]

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बालभारती Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
पाठ 6 Trigonometry
Problem Set 6 | Q 5.02 | पृष्ठ १३८

संबंधित प्रश्‍न

If \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tan​θ.


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


If tanθ = 1 then, find the value of

`(sinθ + cosθ)/(secθ + cosecθ)`


Prove that:

cos2θ (1 + tan2θ)


Prove that: `1/"sec θ − tan θ" = "sec θ + tan θ"`


Prove that:

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

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


Prove the following.

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

Prove the following.

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


Prove the following.

\[\frac{\tan^3 \theta - 1}{\tan\theta - 1} = \sec^2 \theta + \tan\theta\]

Show that: 

`sqrt((1-cos"A")/(1+cos"A"))=cos"ecA - cotA"`


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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