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

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

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प्रश्न

Prove that:

cos2θ (1 + tan2θ)

बेरीज

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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Application of Trigonometry
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Trigonometry - Practice Set 6.1 [पृष्ठ १३१]

संबंधित प्रश्‍न

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:

\[\cot\theta + \tan\theta = cosec\theta \sec\theta\]

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


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.

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 (1 – sinθ) (secθ + tanθ) = 1


Prove the following.
cot2θ – tan2θ = cosec2θ – sec2θ


Prove the following:

sec6x – tan6x = 1 + 3sec2x × tan2x


Choose the correct alternative: 
sinθ × cosecθ =?


Show that: 

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


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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