English

Prove the following.cot2θ – tan2θ = cosec2θ – sec2θ - Geometry Mathematics 2

Advertisements
Advertisements

Question

Prove the following.
cot2θ – tan2θ = cosec2θ – sec2θ

Sum

Solution

L.H.S = \[\cot^2 \theta - \tan^2 \theta\]

[1 + tan2θ = sec2θ, 1 + cot2θ = coses2θ]

\[ = \left( {cosec}^2 \theta - 1 \right) - \left( \sec^2 \theta - 1 \right)\]

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

\[ = {cosec}^2 \theta - \sec^2 \theta\]

= R.H.S

shaalaa.com
Application of Trigonometry
  Is there an error in this question or solution?
Chapter 6: Trigonometry - Problem Set 6 [Page 138]

APPEARS IN

RELATED QUESTIONS

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: `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.
sin \[\theta\] cosec \[\theta\]= ?


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


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.

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

Choose the correct alternative: 
sinθ × cosecθ =?


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×