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प्रश्न
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
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उत्तर
LHS = `(sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta)`
=` ((sin theta + cos theta )^2 + ( sin theta - cos theta)^2)/(( sin theta - cos theta ) ( sin theta + cos theta))`
=`( sin^2 theta + cos^2 theta + 2 sin theta cos theta + sin^2 theta + cos^2 theta - 2 sin theta cos theta)/((sin^2 theta - cos^2 theta))`
=`(1+1)/((- cos^ 2theta )- cos^2 theta) (∵ sin^ 2theta + cos^2 theta =1)`
=`2/(1-2 cos^2 theta)`
= RHS
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संबंधित प्रश्न
If secθ + tanθ = p, show that `(p^{2}-1)/(p^{2}+1)=\sin \theta`
Prove that: `(1 – sinθ + cosθ)^2 = 2(1 + cosθ)(1 – sinθ)`
Prove the following trigonometric identity:
`sqrt((1 + sin A)/(1 - sin A)) = sec A + tan A`
Show that none of the following is an identity:
`sin^2 theta + sin theta =2`
If `cos theta = 7/25 , "write the value of" ( tan theta + cot theta).`
If 3 `cot theta = 4 , "write the value of" ((2 cos theta - sin theta))/(( 4 cos theta - sin theta))`
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
Find the value of sin 30° + cos 60°.
Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2.
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
Prove that `sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A - 1) = 1`.
Prove the following identities:
`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.
Prove the following identities.
`(cot theta - cos theta)/(cot theta + cos theta) = ("cosec" theta - 1)/("cosec" theta + 1)`
The value of sin2θ + `1/(1 + tan^2 theta)` is equal to
If cosA + cos2A = 1, then sin2A + sin4A = 1.
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
Eliminate θ if x = r cosθ and y = r sinθ.
