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प्रश्न
`(sec theta + tan theta )/( sec theta - tan theta ) = ( sec theta + tan theta )^2 = 1+2 tan^2 theta + 25 sec theta tan theta `
उत्तर
Here, `(sec theta + tan theta ) /( sec theta - tan theta)`
=`((sec theta + tan theta ) ( sec theta + tan theta))/(( sec theta - tan theta ) ( sec theta + tan theta ))`
=` ((sec theta + tan theta )^2) /( sec^2 theta - tan^2 theta)`
=`((sec theta + tan theta )^2)/1`
=`(sec theta + tan theta )^2`
Again , `(sec theta + tan theta )2`
=` sec^2 theta + tan^2 theta + 2 sec theta tan theta `
=` 1+ tan^2 theta + tan^2 theta + 2 sec theta tan theta`
=`1+2 tan^2 theta + 2 sec theta tan theta `
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संबंधित प्रश्न
Prove the following trigonometric identities.
`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`
`sin^2 theta + cos^4 theta = cos^2 theta + sin^4 theta`
`sqrt((1-cos theta)/(1+cos theta)) = (cosec theta - cot theta)`
`(sin theta+1-cos theta)/(cos theta-1+sin theta) = (1+ sin theta)/(cos theta)`
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`
If sinθ = `11/61`, find the values of cosθ using trigonometric identity.
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
If \[\cos A = \frac{7}{25}\] find the value of tan A + cot A.
\[\frac{1 - \sin \theta}{\cos \theta}\] is equal to
Prove the following identity :
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
Without using trigonometric identity , show that :
`tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ = 1/sqrt(3)`
Prove that : `1 - (cos^2 θ)/(1 + sin θ) = sin θ`.
If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= R.H.S
