English
Maharashtra State BoardSSC (English Medium) 10th Standard

If tan ⁡θ = 7/24, then to find value of cos θ complete the activity given below. Activity: sec^2θ = 1 + □ ...[Fundamental tri. identity] sec^2θ = 1 + □^2 sec^2θ = 1 + □/576 sec^2θ = □/576 sec θ = □

Advertisements
Advertisements

Question

If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.

Activity:

sec2θ = 1 + `square`   ...[Fundamental tri. identity]

sec2θ = 1 + `square^2`

sec2θ = 1 + `square/576`

sec2θ = `square/576`

sec θ = `square` 

cos θ = `square`   ...`[cos theta = 1/sectheta]`

Fill in the Blanks
Sum
Advertisements

Solution

\[\text{sec}^2θ = 1 + \boxed{\text{tan}^2θ}\]   ...[Fundamental tri. identity]

∴ \[\text{sec}^2θ = 1 + \boxed{\frac{7}{24}}^2\]

∴ \[\text{sec}^2θ = 1 + \frac{\boxed{49}}{576}\]

∴ sec2θ = `(576 + 49)/576`

∴ \[\text{sec}^2θ = \frac{\boxed{625}}{576}\]

∴ \[\text{sec} \phantom{.}θ = \boxed{\frac{25}{24}}\]

∴ \[\text{cos} \phantom{.}θ = \boxed{\frac{24}{25}}\]   ...`[cos theta = 1/sectheta]`

shaalaa.com
  Is there an error in this question or solution?
Chapter 6: Trigonometry - Exercise

RELATED QUESTIONS

 Evaluate sin25° cos65° + cos25° sin65°


Show that `sqrt((1+cosA)/(1-cosA)) = cosec A + cot A`


Prove the following trigonometric identity.

`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`


Prove the following identities:

`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`


`(1 + cot^2 theta ) sin^2 theta =1`


` tan^2 theta - 1/( cos^2 theta )=-1`


Prove the following identities:

`("cosec"  theta + cot theta)/("cosec"  theta - cot theta) = ("cosec"  theta + cot theta )^2 = 1 + 2 cot^2 theta + 2  "cosec"  theta cot theta`


Prove the following identities:

`(1 + tan theta + cot theta)(sin theta - cos theta) = ((sec theta)/("cosec"^2 theta) - ("cosec"  theta)/(sec^2 theta))`


Write the value of tan1° tan 2°   ........ tan 89° .


Prove the following identity :

`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`


Prove the following identity : 

`sin^8θ - cos^8θ = (sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`


Prove the following identity :

`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`


Prove the following identity : 

`[1/((sec^2θ - cos^2θ)) + 1/((cosec^2θ - sin^2θ))](sin^2θcos^2θ) = (1 - sin^2θcos^2θ)/(2 + sin^2θcos^2θ)`


Prove that:

tan (55° + x) = cot (35° – x)


Prove that `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1) = (1 + cos "A")/sin "A"`


Prove the following identities: sec2 θ + cosec2 θ = sec2 θ cosec2 θ.


Prove the following identities.

tan4 θ + tan2 θ = sec4 θ – sec2 θ


If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to 


If sec θ = `25/7`, find the value of tan θ.

Solution:

1 + tan2 θ = sec2 θ

∴ 1 + tan2 θ = `(25/7)^square`

∴ tan2 θ = `625/49 - square`

= `(625 - 49)/49`

= `square/49`

∴ tan θ = `square/7` ........(by taking square roots)


Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×