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प्रश्न
Without using trigonometric identity , show that :
`tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ = 1/sqrt(3)`
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उत्तर
`tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ = 1/sqrt(3)`
Consider `tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ`
⇒ `tan(90^circ - 80^circ) - tan(90^circ - 70^circ) tan30^circ tan70^circ tan80^circ`
⇒ `cot80^circ . cot70^circ .tan30^circ tan70^circ tan80^circ`
⇒ `tan30^circ = 1/sqrt(3)` [As tanθ cotθ = 1]
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(cot A - cos A)/(cot A + cos A) = (cosec A - 1)/(cosec A + 1)`
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Prove the following identities:
`1/(1 + cosA) + 1/(1 - cosA) = 2cosec^2A`
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
`(1+ cos theta + sin theta)/( 1+ cos theta - sin theta )= (1+ sin theta )/(cos theta)`
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Choose the correct alternative:
1 + tan2 θ = ?
Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)
