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If `Sec Theta = X ,"Write the Value of Tan" Theta`. - Mathematics

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प्रश्न

If `sec theta = x ,"write the value of tan"  theta`.

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उत्तर

As , `tan^2 theta = sec^2 theta -1 `

So, `tan theta = sqrt( sec^2 theta -1 ) = sqrt( x^2 -1)`

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अध्याय 8: Trigonometric Identities - Exercises 3

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आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 8 Trigonometric Identities
Exercises 3 | Q 40

संबंधित प्रश्न

Prove the following trigonometric identities.

`(cos^2 theta)/sin theta - cosec theta +  sin theta  = 0`


Prove the following trigonometric identities.

if `T_n = sin^n theta + cos^n theta`, prove that `(T_3 - T_5)/T_1 = (T_5 - T_7)/T_3`


Prove the following trigonometric identities.

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


Prove that:

`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`


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


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


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


If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`


Define an identity.


Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\] 


What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]


Prove the following identity :

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


Prove the following identity : 

`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`


Prove the following identity : 

`((1 + tan^2A)cotA)/(cosec^2A) = tanA`


Prove the following identity : 

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


Prove the following identity : 

`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`


Prove the following identity :

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


Find the value of x , if `cosx = cos60^circ cos30^circ - sin60^circ sin30^circ`


Prove that `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2 


Prove that

sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`


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