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प्रश्न
If tanθ `= 3/4` then find the value of secθ.
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उत्तर
If tanθ = 34
1 + tan2θ = sec2θ
∴ 1 + `(3/4)^2= sec^2θ`
∴ `1 + 9/16 = sec^2θ`
∴ `25/16 = sec^2θ`
∴ `secθ = 5/4`
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संबंधित प्रश्न
Prove the following trigonometric identities.
`sqrt((1 - cos theta)/(1 + cos theta)) = cosec theta - cot theta`
Prove the following trigonometric identities.
`(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)
Show that one of the values of each member of this equality is sin α sin β sin γ
Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`
`(sec^2 theta-1) cot ^2 theta=1`
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
` (sin theta - cos theta) / ( sin theta + cos theta ) + ( sin theta + cos theta ) / ( sin theta - cos theta ) = 2/ ((2 sin^2 theta -1))`
`(1+ cos theta - sin^2 theta )/(sin theta (1+ cos theta))= cot theta`
Write the value of `( 1- sin ^2 theta ) sec^2 theta.`
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
Prove the following identity :
`sin^8θ - cos^8θ = (sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`
Find the value of x , if `cosx = cos60^circ cos30^circ - sin60^circ sin30^circ`
Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.
Prove that: sin4 θ + cos4θ = 1 - 2sin2θ cos2 θ.
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
If 3 sin A + 5 cos A = 5, then show that 5 sin A – 3 cos A = ± 3
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
