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प्रश्न
If A = 60°, B = 30° verify that tan( A - B) = `(tan A - tan B)/(1 + tan A. tan B)`.
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उत्तर
It is given that A = 60°, B = 30°
Putting A = 60° and B = 30° in the given equation,
we get
tan( A - B) = `(tan A - tan B)/(1 + tan A. tan B)`
⇒ tan( 60° - 30° ) = `(tan 60° - tan 30° )/(1 + tan 60°. tan 30° )`
⇒ tan 30° = `(sqrt3 - 1/sqrt3)/(1 + sqrt3 xx 1/sqrt3)`
⇒ `((3-1)/sqrt3)/2`
⇒ `(2/sqrt3)/(2/1)`
⇒ `2/(2sqrt3)`
⇒ `1/sqrt3`
⇒ LHS = RHS
संबंधित प्रश्न
If x cos A + y sin A = m and x sin A – y cos A = n, then prove that : x2 + y2 = m2 + n2
Prove the following identities:
`1/(cosA + sinA) + 1/(cosA - sinA) = (2cosA)/(2cos^2A - 1)`
Write the value of ` sec^2 theta ( 1+ sintheta )(1- sintheta).`
If `cosec theta = 2x and cot theta = 2/x ," find the value of" 2 ( x^2 - 1/ (x^2))`
What is the value of (1 − cos2 θ) cosec2 θ?
If tan α = n tan β, sin α = m sin β, prove that cos2 α = `(m^2 - 1)/(n^2 - 1)`.
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
Prove the following:
(sin α + cos α)(tan α + cot α) = sec α + cosec α
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 sinθ = `11/61`, then find the value of cosθ using the trigonometric identity.
