Advertisements
Advertisements
प्रश्न
If `secθ = 25/7 ` then find tanθ.
Advertisements
उत्तर
`1 + tan^2θ = sec^2θ`
`1 + tan^2θ =(25/7)^2`
`∴ tan^2θ =625/49- 1`
`∴ tan^2θ =(625-49)/49`
`∴ tan^2θ =576/49`
`∴ tanθ =24/7`
APPEARS IN
संबंधित प्रश्न
Prove that `\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }+\frac{\sin\theta +\cos \theta }{\sin \theta -\cos \theta }=\frac{2}{2\sin^{2}\theta -1}`
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`
Prove the following trigonometric identities.
`sin^2 A + 1/(1 + tan^2 A) = 1`
Prove the following trigonometric identities
`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
Prove the following identities:
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove the following identities:
`(sintheta - 2sin^3theta)/(2cos^3theta - costheta) = tantheta`
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
`(1+ cos theta)(1- costheta )(1+cos^2 theta)=1`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
If cos (\[\alpha + \beta\]= 0 , then sin \[\left( \alpha - \beta \right)\] can be reduced to
Prove the following identity :
`(cosecA - sinA)(secA - cosA)(tanA + cotA) = 1`
Prove the following identity :
`(cotA + tanB)/(cotB + tanA) = cotAtanB`
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
If sec θ = `25/7`, then find the value of tan θ.
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.
Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`
Prove that `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2
