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

Prove that (cot A)/(1 – tan A) + (tan A)/(1 – cot A) = 1 + tan A + cot A = sec A . cosec A + 1.

Advertisements
Advertisements

प्रश्न

Prove that `(cot A)/(1 - tan A) + (tan A)/(1 - cot A) = 1 + tan A + cot A = sec A  .  "cosec"  A + 1`.

सिद्धांत
Advertisements

उत्तर

`(cot A)/(1 - tan A) + (tan A)/(1 - cot A)`

= `((cos A)/(sin A))/(1 - (sin A)/(cos A)) + ((sin A)/(cos A))/(1 - (cos A)/(sin A))`

= `((cos A)/(sin A))/((cos A  -  sin A)/(cos A)) + ((sin A)/(cos A))/((sin A  -  cos A)/(sin A))`

= `(cos A)/(sin A) xx (cos A)/(cos A - sin A) + (sin A)/(cos A) xx (sin A)/(sin A - cos A)`

= `(cos^2A)/(sin A(cos A - sin A)) + (sin^2A)/(cos A(sin A - cos A))`

= `1/(sin A - cos A) ((-cos^3A + sin^3A)/(sin A cos A))`

= `1/(sin A - cos A)((sin^3A - cos^3A)/(sin A cos A))`

= `1/(sin A - cos A) xx ((sin A - cos A)(sin^2A + sin A cos A + cos^2A))/(sin A cos A)`   ...[∵ a3 – b3 = (a – b)(a2 + ab + b2)]

= `(sin^2A + sin A cos A + cos^2A)/(sin A cos A)`   ...(i)

= `(1 + sin A cos A)/(sin A cos A)`   ...[∵ sin2A + cos2A = 1]

= `1/(sin A cos A) + (sin A cos A)/(sin A cos A)`

= cosec A sec A + 1   ...(ii)

`(cot A)/(1 - tan A) + (tan A)/(1 - cot A)`

= `(sin^2A + sin A cos A + cos^2A)/(sin A cos A)`   ...[From (i)]

= `(sin^2A)/(sin A cos A) + (sin A cos A)/(sin A cos A) + (cos^2A)/(sin A cos A)`

= `(sin A)/(cos A) + 1 + (cos A)/(sin A)`

= tan A + 1 + cot A   ...(iii)

From (ii) and (iii), we get

`(cot A)/(1 - tan A) + (tan A)/(1 - cot A) = 1 + tan A + cot A = sec A  .  "cosec"  A + 1`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Trigonometry - Exercise

संबंधित प्रश्‍न

Prove the following trigonometric identities.

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


Prove the following trigonometric identities.

if x = a cos^3 theta, y = b sin^3 theta` " prove that " `(x/a)^(2/3) + (y/b)^(2/3) = 1`


Prove the following identities:

`1 - cos^2A/(1 + sinA) = sinA`


Show that : tan 10° tan 15° tan 75° tan 80° = 1


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


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


`(sec theta -1 )/( sec theta +1) = ( sin ^2 theta)/( (1+ cos theta )^2)`


` (sin theta - cos theta) / ( sin theta + cos theta ) + ( sin theta + cos theta ) / ( sin theta - cos theta ) = 2/ ((2 sin^2 theta -1))`


Prove the following identities:

`(cot^2 theta (sec theta - 1))/((1 + sin theta)) + (sec^2 theta(sin theta - 1))/((1 + sec theta)) = 0`


If (cot θ + tan θ) = m and (sec θ – cos θ) = n, prove that `(m^2 n)^(2//3) - (mn^2)^(2//3) = 1`.


If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`


\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to 

 

 


Prove the following identity : 

`sec^4A - sec^2A = sin^2A/cos^4A`


Prove the following identity : 

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


Evaluate:

`(tan 65^circ)/(cot 25^circ)`


Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`


Prove the following identities:

`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.


The value of sin2θ + `1/(1 + tan^2 theta)` is equal to 


Prove that `(1 + sin B)/(cos B) + (cos B)/(1 + sin B) = 2 sec B`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×