Question

If 7 sin2θ+ cos2θ=44, show that tan=1/√3

Answer 1

Step-by-step explanation:

Given Question:-

If 7 sin2θ+ cos2θ=44, show that tan=1/√3

Correct Question:-

If 7 sin^2θ+ 3cos^2θ=4, show that

tan θ=1/√3

Solution:-

Given that :

7 Sin^2 θ + 3Cos^2 θ = 4

It can be written as

4Sin^2 θ + 3Sin^2 θ+ 3 Cos^2 θ = 4

=>4 Sin^2 θ +( 3 Sin^2 θ+3 Cos^2 θ) = 4

=>4 Sin^2 θ + 3( Sin^2 θ+ Cos^2 θ) = 4

we know that Sin^2 θ + Cos^2 θ = 1

=>4 Sin^2 θ + 3(1) = 4

=> 4Sin^2 θ + 3 = 4

=>4 Sin^2 θ = 4-3

=> 4Sin^2 θ = 1

=>Sin^2 θ = 1/4

=> Sin θ = √(1/4)

=>Sin θ = 1/2

=>Sin θ = Sin 30°

=> θ = 30°

The value of θ = 30°

Now , Tan θ

=>tan 30°

=>1/√3

Tan θ = 1/√3

Hence proved

Answer:-

If 7 sin^2θ+ 3cos^2θ=4, then tanθ =1/√3

Check:-

LHS:-

7 sin^2θ+ 3cos^2θ=4

If θ= 30° then

7 sin^2(30°) + 3 Cos^2(30°)

=>7(1/2)^2 + 3(√3/2)^2

=>7(1/4) + 3(3/4)

=>(7/4) + (9/4)

=>(7+9)/4

=>16/4

=>4

RHS

LHS = RHS is true for θ= 30°

Used formulae:-

• Sin^2 θ + Cos^2 θ = 1
• Sin 30° = 1/2
• Tan θ = 1/√3
• Cos 30° = √3/2

Answer 2

Notice,

7sin2θ+3cos2θ=4

7tan2θsec2θ+3sec2θ=4

7tan2θ+3sec2θ=4

7tan2θ+3=4sec2θ

7tan2θ+3=4(1+tan2θ)

7tan2θ+3=4+4tan2θ

3tan2θ=1

tan2θ=13

tanθ=±13√