Question

6. If tana = √2-1 then the sina cosa = ____​

Answer 1

Step-by-step explanation:

Given:-

Tan A =√2-1

To find:-

Find the value of SinA CosA ?

Solution:-

Given that

Tan A =√2-1

On squaring both sides then

=>Tan^2 A = (√2-1)^2

We know that

(a-b)^2=a^2-2ab+b^2

=>Tan^2 A= (√2)^2 -2(√2)(1)+(1)^2

=>Tan^2 A=2-2√2+1

=>Tan^2A=3-2√2

We know that

Sec^2 A- Tan^2 A=1

=>Sec^2 A=1+Tan^2 A

=>1+Tan^2 A = 1+3-2√2

=>Sec^2 A = 4-2√2

=>Sec A =√[4-2√2]

=>1/Cos A = √(4-2√2)

=> Cos A =1/√(4-2√2)-----(1)

Now on squaring both sides

Cos^2 A = 1/(4-2√2)

We know that

Sin^2 A + Cos^2 A = 1

=> Sin^2 A = 1-Cos^2 A

=> 1- Cos^2 A = 1-[1/(4-2√2)]

=> Sin^2 A = [(4-2√2)-1]/(4-2√2)

=>Sin^2 A = (3-2√2)/(4-2√2)

=> Sin A =√(3-2√2)/√(4-2√2)-----(2)

Now Sin A. Cos A

From (1)&(2)

Sin A.Cos A

=>√(3-2√2)/√(4-2√2)× 1/√(4-2√2)

=> √(3-2√2)/√[(4-2√2)(4-2√2)]

=>√(3-2√2)/(4-2√2)

It can be written as

=>√(3-2√2)(4+2√2)/(4-2√2)(4+2√2)

=>√(√2-1)^2(4+2√2)/[(4)^2-(2√2)^2]

=>(√2-1)(4+2√2)/(16-8)

=>(4√2+4-4-2√2)/8

=>(4√2-2√2)/8

=>2√2/8

=>√2/4

SinA CosA = √2/4

Answer:-

The value of SinA CosA for the given problem is

√2/4

Used formulae:-

• (a-b)^2=a^2-2ab+b^2

• Sin^2 A + Cos^2 A = 1

• Sec^2 A- Tan^2 A=1