Advertisements
Advertisements
प्रश्न
Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`
Advertisements
उत्तर
`(1+ tan^2 theta )(1+ sin theta )(1- sintheta)`
=` sec^2 theta (1- sin^2 theta )`
=`1/ cos^2 theta xx cos^2 theta`
= 1
APPEARS IN
संबंधित प्रश्न
`"If "\frac{\cos \alpha }{\cos \beta }=m\text{ and }\frac{\cos \alpha }{\sin \beta }=n " show that " (m^2 + n^2 ) cos^2 β = n^2`
Prove the following trigonometric identities.
`(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`
Prove the following trigonometric identities.
`tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`
Prove the following trigonometric identities.
`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`
Prove the following trigonometric identities.
(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
Prove the following trigonometric identities.
tan2 A sec2 B − sec2 A tan2 B = tan2 A − tan2 B
Prove the following identities:
`1/(secA + tanA) = secA - tanA`
Prove the following identities:
`cosecA + cotA = 1/(cosecA - cotA)`
Prove the following identities:
`cosA/(1 - sinA) = sec A + tan A`
Prove that:
`1/(cosA + sinA - 1) + 1/(cosA + sinA + 1) = cosecA + secA`
Prove the following identities:
`cosecA - cotA = sinA/(1 + cosA)`
`sin^6 theta + cos^6 theta =1 -3 sin^2 theta cos^2 theta`
`(sec theta + tan theta )/( sec theta - tan theta ) = ( sec theta + tan theta )^2 = 1+2 tan^2 theta + 25 sec theta tan theta `
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
Prove the following identity :
`cosec^4A - cosec^2A = cot^4A + cot^2A`
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
Prove that:
`sqrt(( secθ - 1)/(secθ + 1)) + sqrt((secθ + 1)/(secθ - 1)) = 2 "cosec"θ`
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
