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प्रश्न
What is the value of \[\frac{\tan^2 \theta - \sec^2 \theta}{\cot^2 \theta - {cosec}^2 \theta}\]
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उत्तर
We have,
\[\frac{\tan^2 \theta - \sec^2 \theta}{\cot^2 \theta - {cosec}^2 \theta}\]=` (-1(sec ^2 θ-tan ^2θ ))/(-1 (cosec^2 θ-cot ^2 θ))`
=`( secx^2θ-tan^2 θ)/ (cosec ^2 θ-cot^2 θ)`
We know that,
`sec^2θ-tan ^2θ=1`
` cosec^2 θ-cot ^2θ=1`
Therefore,
`(tan ^2θ-sec^2 θ)/(cot^2θ-cosec^2 θ)=1/1`
=1
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संबंधित प्रश्न
Prove that ` \frac{\sin \theta -\cos \theta +1}{\sin\theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }` using the identity sec2 θ = 1 + tan2 θ.
Prove the following trigonometric identities.
`(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
Prove the following trigonometric identities.
`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
Prove the following trigonometric identities.
`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
Prove that:
cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
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`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
`cos^2 theta + 1/((1+ cot^2 theta )) =1`
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
`tan theta /((1 - cot theta )) + cot theta /((1 - tan theta)) = (1+ sec theta cosec theta)`
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
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`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
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Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= R.H.S
If cosA + cos2A = 1, then sin2A + sin4A = 1.
If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.
