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The Value of (1 + Cot θ − Cosec θ) (1 + Tan θ + Sec θ) is - Mathematics

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Question

The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is 

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MCQ
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Solution

The given expression is 

`(1+cot θ-cosec θ )(1+tan θ+sec θ)`

Simplifying the given expression, we have 

`(1+cot θ-cosec θ)(1+tan θ+sec θ)` 

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

= `(sin θ+cos θ-1)/sin θxx (cos θ+sin θ+1)/cos θ` 

= `((sin θ+cos θ-1)(cos θ+sin θ+1))/(sin θcos θ)` 

=`({(sin θ+cos θ)-1}{(sin θ+cos θ)+1})/(sin θ cos θ)`

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

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

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

=`((sin ^2θ+cos^2θ)+2 sinθ cos θ-1)/(sin θcos θ)` 

= `(1+2 sin θ cosθ-1)/(sinθ cos θ)` 

=`( 2 sin θ cos θ)/(sin θ cos θ)` 

=`2`

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Chapter 11: Trigonometric Identities - Exercise 11.4 [Page 57]

APPEARS IN

RD Sharma Mathematics [English] Class 10
Chapter 11 Trigonometric Identities
Exercise 11.4 | Q 9 | Page 57

RELATED QUESTIONS

Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`


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


If m = a sec A + b tan A and n = a tan A + b sec A, then prove that : m2 – n2 = a2 – b2


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


If `( cosec theta + cot theta ) =m and ( cosec theta - cot theta ) = n, ` show that mn = 1.


If `sec theta + tan theta = p,` prove that

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 Write True' or False' and justify your answer the following :

The value of the expression \[\sin {80}^° - \cos {80}^°\] 


Prove the following identity :

`(1 - sin^2θ)sec^2θ = 1`


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Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.


Prove that `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ)) = 1/(tan θ + cot θ)`


If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.


tan θ cosec2 θ – tan θ 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)


(sec θ + tan θ) . (sec θ – tan θ) = ?


To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.

Activity:

L.H.S = `square`

= `square/sintheta + sintheta/costheta`

= `(cos^2theta + sin^2theta)/square`

= `1/(sintheta*costheta)`     ......`[cos^2theta + sin^2theta = square]`

= `1/sintheta xx 1/square`

= `square`

= R.H.S


Show that, cotθ + tanθ = cosecθ × secθ

Solution :

L.H.S. = cotθ + tanθ

= `cosθ/sinθ + sinθ/cosθ`

= `(square + square)/(sinθ xx cosθ)`

= `1/(sinθ xx cosθ)` ............... `square`

= `1/sinθ xx 1/square`

= cosecθ × secθ

L.H.S. = R.H.S

∴ cotθ + tanθ = cosecθ × secθ


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