Advertisements
Advertisements
प्रश्न
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
Advertisements
उत्तर
LHS = `( tan A + sec A - 1)/(tan A - sec A + 1)`
= `(( tan A + sec A) - (sec^2 A - tan^2 A))/((tan A - sec A) + 1)`
= `(( tan A + sec A)( 1 - sec A + tan A))/(tan A - sec A + 1)`
= tan A + sec A
= `sin A/cos A + 1/cos A = (1 + sin A)/cos A`
= RHS
Hence proved.
संबंधित प्रश्न
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
`sqrt((1 + sin θ)/(1 - sin θ)) = sec θ + tan θ`
Write the value of ` sin^2 theta cos^2 theta (1+ tan^2 theta ) (1+ cot^2 theta).`
Prove that:
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.
If `tan θ = 13/12`, then cot θ = ?
If `sec θ + tan θ = sqrt(3)`, complete the activity to find the value of sec θ – tan θ.
Activity:
`square = 1 + tan^2θ` ...[Fundamental trigonometric identity]
`square - tan^2θ = 1`
`(sec θ + tan θ) . (sec θ - tan θ) = square`
`sqrt(3) . (sec θ - tan θ) = 1`
`(sec θ - tan θ) = square`
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
Prove that sin4A – cos4A = 1 – 2 cos2A.
