Advertisements
Advertisements
Question
tan θ × `sqrt(1 - sin^2 θ)` is equal to:
Options
cos θ
sin θ
tan θ
cot θ
Solution
sin θ
Explanation:
`tan θ xx sqrt(1 - sin^2 θ) ...{sin^2 θ + cos^2 θ = 1, ∴ cos^2 θ = 1 - sin^2 θ}`
= `tan θ xx sqrt(cos^2 θ)`
= tan θ × cos θ
= `(sin θ)/(cos θ)` × cos θ
= sin θ
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
Prove the following trigonometric identities.
if x = a cos^3 theta, y = b sin^3 theta` " prove that " `(x/a)^(2/3) + (y/b)^(2/3) = 1`
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
If 5x = sec θ and \[\frac{5}{x} = \tan \theta\]find the value of \[5\left( x^2 - \frac{1}{x^2} \right)\]
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
Choose the correct alternative:
`(1 + cot^2"A")/(1 + tan^2"A")` = ?
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
