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`(Tan a + Tanb )/(Cot a + Cot B) = Tan a Tan B` - Mathematics

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`(tan A + tanB )/(cot A + cot B) = tan A tan B`

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LHS = `(tan A + tanB )/(cot A + cot B) `

       =`(tan A + tan B)/(1/ tan A + 1/ tanB)`

       =` (tan A + tan B)/( (tan A+tan B)/ (tan A tan B)`

        =`(tan A tan B ( tan A + tan B))/((tan A + tan B ))`

        = ЁЭСбЁЭСОЁЭСЫЁЭР┤ ЁЭСбЁЭСОЁЭСЫЁЭР╡
        = RHS
Hence, LHS = RHS

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рдЕрдзреНрдпрд╛рдп 8: Trigonometric Identities - Exercises 1

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рдЖрд░рдПрд╕ рдЕрдЧреНрд░рд╡рд╛рд▓ Mathematics [English] Class 10
рдЕрдзреНрдпрд╛рдп 8 Trigonometric Identities
Exercises 1 | Q 35

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If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p2 – 1) = 2p


Prove the following trigonometric identities.

(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1


Prove the following trigonometric identities

If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 − y2 = a2 − b2


Prove the following identities:

`(secA - tanA)/(secA + tanA) = 1 - 2secAtanA + 2tan^2A`


Prove that:

`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`


If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that : x2 + y2 + z2 = r2


`sin theta / ((1+costheta))+((1+costheta))/sin theta=2cosectheta` 


Find the value of ` ( sin 50°)/(cos 40°)+ (cosec 40°)/(sec 50°) - 4 cos 50°   cosec 40 °`


Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`


(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to


If cos A + cos2 A = 1, then sin2 A + sin4 A =


Prove the following identity :

`sec^2A.cosec^2A = tan^2A + cot^2A + 2`


Prove the following identity : 

`sec^4A - sec^2A = sin^2A/cos^4A`


Prove the following identity : 

`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`


Find the value of `θ(0^circ < θ < 90^circ)` if : 

`tan35^circ cot(90^circ - θ) = 1`


If A = 60°, B = 30° verify that tan( A - B) = `(tan A - tan B)/(1 + tan A. tan B)`.


If sec θ = `25/7`, find the value of tan θ.

Solution:

1 + tan2 θ = sec2 θ

∴ 1 + tan2 θ = `(25/7)^square`

∴ tan2 θ = `625/49 - square`

= `(625 - 49)/49`

= `square/49`

∴ tan θ = `square/7` ........(by taking square roots)


If 1 – cos2θ = `1/4`, then θ = ?


Prove that (1 – cos2A) . sec2B + tan2B(1 – sin2A) = sin2A + tan2B


Prove the following:

`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ


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