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Questions
Prove the following trigonometric identities.
`"cosec" theta sqrt(1 - cos^2 theta) = 1`
Prove the following:
`"cosec" theta sqrt(1 - cos^2 theta) = 1`
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Solution
We know that `sin^2 theta + cos^2 theta = 1`
So,
LHS = `"cosec" theta sqrt(1 - cos^2 theta)`
= `"cosec" theta sqrt (sin^2 theta)`
= cosec θ . sin θ
`1/sin theta xx sin theta`
= 1
= RHS hence proved.
RELATED QUESTIONS
Prove the following identities:
`((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA)) = 2cotA`
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
If `sec theta = x ,"write the value of tan" theta`.
Prove that:
Sin4θ - cos4θ = 1 - 2cos2θ
If `secθ = 25/7 ` then find tanθ.
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
Without using the trigonometric table, prove that
tan 10° tan 15° tan 75° tan 80° = 1
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]`
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 `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
