Advertisements
Advertisements
प्रश्न
Prove that sec2θ – cos2θ = tan2θ + sin2θ
उत्तर
L.H.S = sec2θ – cos2θ
= sec2θ – (1 – sin2θ) ......`[(because sin^2theta + cos^2theta = 1),(therefore 1 - sin^2theta = cos^2theta)]`
= sec2θ – 1 + sin2θ
= tan2θ + sin2θ ......`[(because 1 + tan^2theta = sec^2theta),(therefore tan^2theta = sec^2theta - 1)]`
= R.H.S
∴ sec2θ – cos2θ = tan2θ + sin2θ
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`sin^2 A + 1/(1 + tan^2 A) = 1`
Prove the following trigonometric identities.
`(1 + sin theta)/cos theta + cos theta/(1 + sin theta) = 2 sec theta`
`Prove the following trigonometric identities.
`(sec A - tan A)^2 = (1 - sin A)/(1 + sin A)`
Prove the following identities:
cosec A(1 + cos A) (cosec A – cot A) = 1
Prove the following identities:
`secA/(secA + 1) + secA/(secA - 1) = 2cosec^2A`
Prove that:
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
If `(cosec theta - sin theta )= a^3 and (sec theta - cos theta ) = b^3 , " prove that " a^2 b^2 ( a^2+ b^2 ) =1`
If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`
Find the value of ` ( sin 50°)/(cos 40°)+ (cosec 40°)/(sec 50°) - 4 cos 50° cosec 40 °`
Prove the following identity :
`(1 - sin^2θ)sec^2θ = 1`
Prove the following identity :
tanA+cotA=secAcosecA
Prove the following identity :
cosecθ(1 + cosθ)(cosecθ - cotθ) = 1
If tan θ = 2, where θ is an acute angle, find the value of cos θ.
Prove that `tan^3 θ/( 1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ) = sec θ. cosec θ - 2 sin θ cos θ.`
Without using the trigonometric table, prove that
tan 10° tan 15° tan 75° tan 80° = 1
Prove that: `cos^2 A + 1/(1 + cot^2 A) = 1`.
Choose the correct alternative:
sec2θ – tan2θ =?
Prove that sec2θ − cos2θ = tan2θ + sin2θ
Prove that `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2
