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What is the Value of Sin 2 θ + 1 1 + Tan 2 θ - Mathematics

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प्रश्न

What is the value of \[\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\]

थोडक्यात उत्तर
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उत्तर

We have, 

`sin^2 θ+1/(1+tan^2θ)= sin^2θ+1/(sqc^2θ)` 

=` sin^2θ+(1/secθ)^2`  

=` sin^2 θ+cos^2θ` 

=` 1`

 

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पाठ 11: Trigonometric Identities - Exercise 11.3 [पृष्ठ ५५]

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आरडी शर्मा Mathematics [English] Class 10
पाठ 11 Trigonometric Identities
Exercise 11.3 | Q 4 | पृष्ठ ५५

संबंधित प्रश्‍न

Show that `sqrt((1-cos A)/(1 + cos A)) = sinA/(1 + cosA)`


Prove the following trigonometric identities.

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


Prove the following identities:

(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1


Prove the following identities:

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


Prove the following identities:

`cosecA - cotA = sinA/(1 + cosA)`


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


`cot theta/((cosec  theta + 1) )+ ((cosec  theta +1 ))/ cot theta = 2 sec theta `


Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`


If `sec theta + tan theta = x,"  find the value of " sec theta`


Prove the following identity :

 ( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ) 


Prove the following identity :

secA(1 - sinA)(secA + tanA) = 1


If tan θ = 2, where θ is an acute angle, find the value of cos θ. 


Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tanθ + cotθ. 


If sec θ = x + `1/(4"x"), x ≠ 0,` find (sec θ + tan θ)


Prove that `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ)) = 1`.


If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2


Prove that `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ


Prove the following:

`1 + (cot^2 alpha)/(1 + "cosec"  alpha)` = cosec α


Prove the following:

(sin α + cos α)(tan α + cot α) = sec α + cosec α


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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