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Prove the Following.(Secθ + Tanθ) (1 – Sinθ) = Cosθ - Geometry Mathematics 2

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प्रश्न

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

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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Application of Trigonometry
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Trigonometry - Problem Set 6 [पृष्ठ १३८]

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बालभारती Geometry (Mathematics 2) [English] 10 Standard SSC 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 \[\cot\theta = \frac{40}{9}\], find the values of cosecθ and sinθ.


If tanθ = 1 then, find the value of

`(sinθ + cosθ)/(secθ + cosecθ)`


Prove that:

cos2θ (1 + tan2θ)


Prove that:

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


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


Prove that:

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

Prove that:
If \[\tan\theta + \frac{1}{\tan\theta} = 2\], then show that \[\tan^2 \theta + \frac{1}{\tan^2 \theta} = 2\]


Choose the correct alternative answer for the following question.

1 + tan2 \[\theta\]  = ?


Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ 


Prove the following.
cot2θ – tan2θ = cosec2θ – sec2θ


Choose the correct alternative: 
sinθ × cosecθ =?


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.


ΔAMT∼ΔAHE, construct Δ AMT such that MA = 6.3 cm, ∠MAT=120°, AT = 4.9 cm and `"MA"/"HA"=7/5`, then construct ΔAHE.


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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