हिंदी

Choose the Correct Alternative Answer for the Following Question. 1 + Tan2 θ = ?

Advertisements
Advertisements

प्रश्न

Choose the correct alternative answer for the following question.

1 + tan2 \[\theta\]  = ?

विकल्प

  • cot2θ 

  • cosec2θ 

  • sec2θ    

  • tan2θ

MCQ
Advertisements

उत्तर

\[1 + \tan^2 \theta = \sec^2 \theta\]
Hence, the correct answer is sec2θ . 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Trigonometry - Problem Set 6 [पृष्ठ १३८]

APPEARS IN

बालभारती Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
अध्याय 6 Trigonometry
Problem Set 6 | Q 1.3 | पृष्ठ १३८

संबंधित प्रश्न

If \[\cot\theta = \frac{40}{9}\], find the values of cosecθ and sinθ.


Prove that:

cos2θ (1 + tan2θ)


Prove that:

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

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 \[\theta\]= ?


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


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.
(secθ + tanθ) (1 – sinθ) = cosθ


Prove the following.

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

Prove the following:

sec6x – tan6x = 1 + 3sec2x × tan2x


Prove the following.

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


Choose the correct alternative: 
sinθ × cosecθ =?


If sinθ = `8/17`, where θ is an acute angle, find the value of cos θ by using identities.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×