Advertisements
Advertisements
प्रश्न
What is the value of (1 − cos2 θ) cosec2 θ?
उत्तर
We have,
`(1-cos^2 θ) cosec ^2 θ=sin^2θ xx cosec^2 θ`
=` sin^2θ xx(1/sin θ)^2`
= `sin^2 θ xx1/( sin^2θ)`
`= 1`
APPEARS IN
संबंधित प्रश्न
Evaluate
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
As observed from the top of an 80 m tall lighthouse, the angles of depression of two ships on the same side of the lighthouse of the horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest meter.
Prove the following trigonometric identities.
`(cos theta)/(cosec theta + 1) + (cos theta)/(cosec theta - 1) = 2 tan theta`
Prove the following trigonometric identities
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 − y2 = a2 − b2
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
If a cos `theta + b sin theta = m and a sin theta - b cos theta = n , "prove that "( m^2 + n^2 ) = ( a^2 + b^2 )`
sec4 A − sec2 A is equal to
(sec A + tan A) (1 − sin A) = ______.
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Prove the following identity :
`sqrt((secq - 1)/(secq + 1)) + sqrt((secq + 1)/(secq - 1))` = 2 cosesq
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
Prove that `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2 = 1`
Prove that: `(1 + cot^2 θ/(1 + cosec θ)) = cosec θ`.
If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ
Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
If cos θ = `24/25`, then sin θ = ?
Prove that sec2θ − cos2θ = tan2θ + sin2θ
Prove that (1 – cos2A) . sec2B + tan2B(1 – sin2A) = sin2A + tan2B