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प्रश्न
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
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उत्तर
`(tan^2 theta - sec^2 theta )/ (cot^2 theta - cosec^2 theta)`
=` (-1)/(-1)`
= 1
APPEARS IN
संबंधित प्रश्न
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
Prove the following trigonometric identities:
(i) (1 – sin2θ) sec2θ = 1
(ii) cos2θ (1 + tan2θ) = 1
Show that `sqrt((1+cosA)/(1-cosA)) = cosec A + cot A`
Prove that `cosA/(1+sinA) + tan A = secA`
Prove the following trigonometric identities.
sec A (1 − sin A) (sec A + tan A) = 1
Prove the following identities:
`cosecA + cotA = 1/(cosecA - cotA)`
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
Prove the following identities:
`(sinA - cosA + 1)/(sinA + cosA - 1) = cosA/(1 - sinA)`
If`( 2 sin theta + 3 cos theta) =2 , " prove that " (3 sin theta - 2 cos theta) = +- 3.`
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
If `secθ = 25/7 ` then find tanθ.
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
prove that `1/(1 + cos(90^circ - A)) + 1/(1 - cos(90^circ - A)) = 2cosec^2(90^circ - A)`
If sin θ = `1/2`, then find the value of θ.
If sin θ + cos θ = `sqrt(3)`, then prove that tan θ + cot θ = 1.
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
Let x1, x2, x3 be the solutions of `tan^-1((2x + 1)/(x + 1)) + tan^-1((2x - 1)/(x - 1))` = 2tan–1(x + 1) where x1 < x2 < x3 then 2x1 + x2 + x32 is equal to ______.
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
