Advertisements
Advertisements
Question
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
Advertisements
Solution
L.H.S. = `(costhetacottheta)/(1 + sintheta)`
= `(costhetacottheta)/(1 + sintheta) xx (1 - sintheta)/(1 - sintheta)`
= `(costhetacottheta(1 - sintheta))/(1 - sin^2theta)`
= `(costheta costheta/sintheta(1 - sintheta))/cos^2theta`
= `(1 - sintheta)/sintheta`
= `1/sintheta - 1`
= cosec θ – 1
APPEARS IN
RELATED QUESTIONS
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
Write the value of tan10° tan 20° tan 70° tan 80° .
If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
If sin θ + sin2 θ = 1 show that: cos2 θ + cos4 θ = 1
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ...[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` ...`[cos theta = 1/sectheta]`
Prove that sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A.
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
