Advertisements
Advertisements
प्रश्न
Prove that : `tan"A"/(1 - cot"A") + cot"A"/(1 - tan"A") = sec"A".cosec"A" + 1`.
Advertisements
उत्तर
LHS = `(sin"A"/cos"A")/(1 - cos"A"/sin"A") + (cos"A"/sin"A")/(1 - sin"A"/cos"A")`
= `(sin"A" sin"A")/(cos"A"(sin"A" - cos"A")) + (cos"A" cos"A")/((cos"A" - sin"A") sin"A"`
= `1/((sin"A" - cos"A")) [(sin^2"A")/cos"A" + (cos^2"A")/(-sin"A")]`
= `(sin^3"A" - cos^3"A")/(sin"A".cos"A"(sin"A" - cos"A"))`
= `((sin"A" - cos"A")(sin^2"A" + cos^2"A" + sin"A". cos"A"))/(sin"A". cos"A"(sin"A" - cos"A")`
= `(1 + sin"A". cos"A")/(sin"A".cos"A")`
= `1/(sin"A".cos"A") + (sin"A".cos"A")/(sin"A".cos"A")`
= `1/sin"A" . 1/cos"A"+ 1`
= sec A.cosec A + 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A`
Prove the following trigonometric identities
sec4 A(1 − sin4 A) − 2 tan2 A = 1
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove the following identities:
cot2 A – cos2 A = cos2 A . cot2 A
`sin theta (1+ tan theta) + cos theta (1+ cot theta) = ( sectheta+ cosec theta)`
If \[\cos A = \frac{7}{25}\] find the value of tan A + cot A.
\[\frac{1 + \tan^2 A}{1 + \cot^2 A}\]is equal to
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
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)
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S. = `square`
= `cos^2θ xx square` ...`[1 + tan^2θ = square]`
= `(cos θ xx square)^2`
= 12
= 1
= R.H.S.
