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प्रश्न
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
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उत्तर
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^2`
∴ tan2 θ = `625/49 - 1`
= `(625 - 49)/49`
= `576/49`
∴ tan θ = `24/7` ........(by taking square roots)
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + sin theta)/cos theta + cos theta/(1 + sin theta) = 2 sec theta`
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`
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`1 - sin^2A/(1 + cosA) = cosA`
Prove the following identities:
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Show that none of the following is an identity:
`sin^2 theta + sin theta =2`
If `( cosec theta + cot theta ) =m and ( cosec theta - cot theta ) = n, ` show that mn = 1.
If 5x = sec ` theta and 5/x = tan theta , " find the value of 5 "( x^2 - 1/( x^2))`
If 5x = sec θ and \[\frac{5}{x} = \tan \theta\]find the value of \[5\left( x^2 - \frac{1}{x^2} \right)\]
Prove the following identity :
`(1 + sinA)/(1 - sinA) = (cosecA + 1)/(cosecA - 1)`
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Evaluate:
sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
If A = 60°, B = 30° verify that tan( A - B) = `(tan A - tan B)/(1 + tan A. tan B)`.
Prove the following identities:
`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.
Prove that the following identities:
Sec A( 1 + sin A)( sec A - tan A) = 1.
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
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`
Prove that
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"`
