Advertisements
Advertisements
Questions
Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Prove the following:
(sec2 θ − 1) (cosec2 θ − 1) = 1
Advertisements
Solution
We know that
sec2 θ − tan2 θ = 1
cosec2 θ − cot2 θ = 1
So,
(sec2 θ − 1)(cosec2 θ − 1) = tan2 θ × cot2 θ
= (tan θ × cot θ)
= `(tan θ xx 1/tan θ)^2`
= (1)2
= 1
RELATED QUESTIONS
Prove the following identities:
`( i)sin^{2}A/cos^{2}A+\cos^{2}A/sin^{2}A=\frac{1}{sin^{2}Acos^{2}A)-2`
`(ii)\frac{cosA}{1tanA}+\sin^{2}A/(sinAcosA)=\sin A\text{}+\cos A`
`( iii)((1+sin\theta )^{2}+(1sin\theta)^{2})/cos^{2}\theta =2( \frac{1+sin^{2}\theta}{1-sin^{2}\theta } )`
Prove the following identities:
`sinA/(1 + cosA) = cosec A - cot A`
Prove the following identities:
`cosA/(1 + sinA) + tanA = secA`
If `m = (cos θ - sin θ)` and `n = (cos θ + sin θ)`, show that `sqrt(m/n) + sqrt(n/m) = 2/sqrt(1 - tan^2θ)`.
Find the value of ` ( sin 50°)/(cos 40°)+ (cosec 40°)/(sec 50°) - 4 cos 50° cosec 40 °`
Prove the following identity :
`((1 + tan^2A)cotA)/(cosec^2A) = tanA`
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
Prove the following identities: sec2 θ + cosec2 θ = sec2 θ cosec2 θ.
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.
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
