Advertisements
Advertisements
प्रश्न
cos θ . sec θ = ?
विकल्प
1
0
`1/2`
`sqrt(2)`
Advertisements
उत्तर
1
Explanation:
`cos θ . sec θ = cos θ . 1/(cos θ)`
= 1
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities:
`(\text{i})\text{ }\frac{\sin \theta }{1-\cos \theta }=\text{cosec}\theta+\cot \theta `
Prove the following trigonometric identities.
`(1 + cos theta - sin^2 theta)/(sin theta (1 + cos theta)) = cot theta`
If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
`1 + (tan^2 θ)/((1 + sec θ)) = sec θ`
If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1
What is the value of (1 + tan2 θ) (1 − sin θ) (1 + sin θ)?
If sec θ + tan θ = x, then sec θ =
Prove the following identity :
`sin^2Acos^2B - cos^2Asin^2B = sin^2A - sin^2B`
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
Find the value of x , if `cosx = cos60^circ cos30^circ - sin60^circ sin30^circ`
Without using trigonometric identity , show that :
`tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ = 1/sqrt(3)`
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
Prove the following identities.
`(sin "A" - sin "B")/(cos "A" + cos "B") + (cos "A" - cos "B")/(sin "A" + sin "B")`
1 + cot2θ = ?
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 `(tan(90 - θ) + cot(90 - θ))/("cosec" θ) = sec θ`.
Prove that `sqrt((1 + cos A)/(1 - cos A)) = "cosec" A + cot A`.
If cosec A – sin A = p and sec A – cos A = q, then prove that `(p^2q)^(2/3) + (pq^2)^(2/3) = 1`.
If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.
sec θ when expressed in term of cot θ, is equal to ______.
