हिंदी

Prove the following. 1 1 − sin θ + 1 1 + sin θ = 2 sec 2 θ - Geometry Mathematics 2

Advertisements
Advertisements

प्रश्न

Prove the following.

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

उत्तर

\[\frac{1}{1 - \sin\theta} + \frac{1}{1 + \sin\theta}\]

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

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

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

\[ = 2 \sec^2 \theta\]

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

APPEARS IN

बालभारती Mathematics 2 [English] Standard 10 Maharashtra State Board
अध्याय 6 Trigonometry
Problem Set 6 | Q 5.06 | पृष्ठ १३८

संबंधित प्रश्न

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


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


If 5 secθ – 12 cosecθ = 0, find the values of secθ, cosθ, and sinθ.


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:

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

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:

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.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×