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प्रश्न
What is the value of (1 + tan2 θ) (1 − sin θ) (1 + sin θ)?
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उत्तर
We have,
`(1+tan^2θ)(1-sinθ)(1+sin θ)=(1+tan ^2 θ){(1-sinθ)(1+sinθ)}`
= `(1+tan^2θ)(1-sin^2θ)`
We know that,
`sec^2θ-tan^2θ=1`
⇒ `sec^2 θ=1+tan^2θ`
`sin^2 θ+cos ^2θ=1`
⇒ `cos^2 θ=1sin^2θ`
Therefore,
`(1+tan^2θ)(1-sin θ)(1+sin θ) = sec^2 θ xxcos^2θ`
= `1/cos^2θ xx cos^2 θ`
=` 1`
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संबंधित प्रश्न
Express the ratios cos A, tan A and sec A in terms of sin A.
Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.
Prove the following trigonometric identities.
tan2θ cos2θ = 1 − cos2θ
Prove the following trigonometric identities.
(1 + cot A − cosec A) (1 + tan A + sec A) = 2
Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)
Show that one of the values of each member of this equality is sin α sin β sin γ
If tan A = n tan B and sin A = m sin B, prove that:
`cos^2A = (m^2 - 1)/(n^2 - 1)`
`sin theta (1+ tan theta) + cos theta (1+ cot theta) = ( sectheta+ cosec theta)`
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
If `( cosec theta + cot theta ) =m and ( cosec theta - cot theta ) = n, ` show that mn = 1.
If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
Prove the following identity:
`cosA/(1 + sinA) = secA - tanA`
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
If x = h + a cos θ, y = k + b sin θ.
Prove that `((x - h)/a)^2 + ((y - k)/b)^2 = 1`.
Prove that: sin4 θ + cos4θ = 1 - 2sin2θ cos2 θ.
Prove the following identities.
`costheta/(1 + sintheta)` = sec θ – tan θ
Choose the correct alternative:
1 + cot2θ = ?
If tan θ = `9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove that `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A
Prove that
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"`
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
