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प्रश्न
9 sec2 A − 9 tan2 A is equal to
विकल्प
1
9
8
0
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उत्तर
Given:
`9 sec^2 A-9 tan^2 A`
`=9 (sec^2 A-tan^2 A)`
We know that, `sec^2 A-tan^2 A=1`
Therefore, `9 sec^2 A-9 tan^2 A=9`
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संबंधित प्रश्न
(secA + tanA) (1 − sinA) = ______.
Prove the following trigonometric identities.
`tan^2 theta - sin^2 theta tan^2 theta sin^2 theta`
Prove the following trigonometric identity:
`sqrt((1 + sin A)/(1 - sin A)) = sec A + tan A`
Prove the following trigonometric identities.
`(cot A - cos A)/(cot A + cos A) = (cosec A - 1)/(cosec A + 1)`
Prove the following trigonometric identities.
`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`
Prove the following identities:
`(cosecA - 1)/(cosecA + 1) = (cosA/(1 + sinA))^2`
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
If tan A = n tan B and sin A = m sin B, prove that:
`cos^2A = (m^2 - 1)/(n^2 - 1)`
`1+(tan^2 theta)/((1+ sec theta))= sec theta`
`tan theta/(1+ tan^2 theta)^2 + cottheta/(1+ cot^2 theta)^2 = sin theta cos theta`
`(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))=0`
If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`
Prove that:
`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.
Prove the following identity :
`(1 + cotA)^2 + (1 - cotA)^2 = 2cosec^2A`
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
If m = a secA + b tanA and n = a tanA + b secA , prove that m2 - n2 = a2 - b2
Evaluate:
sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°
Prove that `( 1 + sin θ)/(1 - sin θ) = 1 + 2 tan θ/cos θ + 2 tan^2 θ` .
If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to
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
