Advertisements
Advertisements
प्रश्न
Write the value of ` cosec^2 (90°- theta ) - tan^2 theta`
उत्तर
`cosec ^2 (90°- theta )- tan^2 theta `
=` sec^2 theta - tan^2 theta`
= 1
APPEARS IN
संबंधित प्रश्न
Prove that (1 + cot θ – cosec θ)(1+ tan θ + sec θ) = 2
Prove the following trigonometric identities.
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
Prove the following identities:
`(1 - 2sin^2A)^2/(cos^4A - sin^4A) = 2cos^2A - 1`
If `cot theta = 1/ sqrt(3) , "write the value of" ((1- cos^2 theta))/((2 -sin^2 theta))`
Prove that:
Sin4θ - cos4θ = 1 - 2cos2θ
\[\frac{1 - \sin \theta}{\cos \theta}\] is equal to
Prove the following identity :
`tan^2A - sin^2A = tan^2A.sin^2A`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
Find A if tan 2A = cot (A-24°).
Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2.
Prove that `( 1 + sin θ)/(1 - sin θ) = 1 + 2 tan θ/cos θ + 2 tan^2 θ` .
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2
If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to
Choose the correct alternative:
cot θ . tan θ = ?
Prove that cot2θ × sec2θ = cot2θ + 1
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Prove that sec2θ – cos2θ = tan2θ + sin2θ
Prove that
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"`
Prove that `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1