Question

Derine tan (x+y) = (tanx + tan y)/
(1- tanx tany)

using this derive tan (x+y+z)​

Answer 1

Question :

Derive tan (x+y) = (tan x + tan y)/(1-tan x*tan y)

ANSWER

Required to derive : -

tan (x+y) = (tan x + tan y)/(1-tan x*tan y)

Formulae used : -

sin (A+B) = sin A*cos B + cos A*sin B

cos (A+B) = cos A*cos B - sin A*sin B

Derivation : -

tan (x+y) = (tan x + tan y)/(1-tan x*tan y)

From LHS part let's derive the RHS part

Consider the LHS part;

tan (x+y)

We know that;

• tan x = (sin x)/(cos x)

[sin (x+y)]/[cos (x+y)]

Using the formulae;

• sin (A+B) = sin A*cos B + cos A*sin B
• cos (A+B) = cos A*cos B - sin A*sin B

[sin x*cos y + cos x*sin y]/[cos x*cos y - sin x*sin y]

Here,

We need to apply a bit of logic !

We need to convert the ratios in numerator & denominator into tan . so, let's divide the numerator & denominator with (cos x*cos y)

This implies;

([sin x*cos y + cos x*sin y]/[cos x*cos y])/([cos x*cos y - sin x*sin y]/[cos x*cos y])

([sin x*cos y]/[cos x*cos y]+[cos x*sin y]/[cos x*cos y])/([cos x*cos y]/[cos x*cos y]-[sin x*sin y]/[cos x*cos y])

([sin x]/[cos x]+[sin y]/[cos y])/(1-[sin x]/[cos x]*[sin y]/[sin y])

Since,

• (sin @)/(cos @) = tan @

(tan x + tan y)/(1-tan x*tan y)

Hence Derived !

Question :

Derive the identity for tan (x+y+z)

ANSWER

Required to derive : -

• The identity for tan (x+y+z)

Formula used : -

tan (A+B) = (tan A + tan B)/(1-tan A*tan B)

Derivation : -

tan (x+y+z)

tan([x+y]+z)

Here,

a = x+y

b = z

Using the formula;

• tan (A+B) = (tan A + tan B)/(1-tan A*tan B)

tan ([x+y]+z) =

(tan [x+y] + tan z)/(1-tan [x+y]*tan z)

Here,

Again we need to apply the formula once again because we have tan (x+y)

This implies;

([tan x + tan y]/[1-tan x*tan y] + tan z)/(1-[tan x + tan y]/[1-tan x*tan y]*tan z)

([tan x + tan y]/[1-tan x*tan y] + tan z)/(1-[tan z(tan x + tan y)]/[1-tan x*tan y])

([tan x + tan y]/[1-tan x*tan y] + tan z)/(1-[tan x*tan z + tan y*tan z]/[1-tan x*tan y])

Taking the LCM respectively in numerator and denominator

([tan x + tan y + tan z{1-tan x*tan y}]/[1-tan x*tan y])/([1{1-tan x*tan y}-{tan x*tan z + tan y*tan z}]/[1-tan x*tan y])

(1-tan x*tan y) get's cancelled in both numerator and denominator

([tan x + tan y + tan z{1-tan x*tan y}]/[1])/([1{1-tan x*tan y}-{tan x*tan z + tan y*tan z}]/[1])

(tan x + tan y + tan z - tan x*tan z*tan y )/(1-tan x*tan y - tan x*tan z - tan y*tan z)

Hence derived !

tan (x+y+z) = (tan x + tan y + tan z - tan x*tan z*tan y )/(1-tan x*tan y - tan x*tan z - tan y*tan z)

Answer 2

Answer:

Solution :

Given tanx = 2/3 --- ( 1 )

tany = 1/5 ----( 2 )

Now ,

tan( x + y )

= [ tan x + tan y ]/( 1 - tanx * tan y )

= [ 2/3 + 1/5 ]/[ 1 - ( 2/3 )( 1/5 ) ]

= ( 10 + 3 )/( 15 - 2 )

= 13/13

= 1

Therefore ,

tan ( x + y ) = tan 45°

=> x + y = 45°

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