Question

If tana=n tanb and sina=m sinb then show that cos²a=m²-1/n²-1

Answer 1

Hey mate .
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☆ I think the question should be

Prove:-- cos^2A=(n^2−1)/(m^2−1)

Then,

sinA=n.sinB ------(1)

tanA=m.tanB

i.e sinA/cosA=m.sinB/cosB

i.e sinA/sinB=m.cosA/cosB

i.e  n.cosB=m.cosA ……..[from (1)]

Squaring both sides gives:

n^2.cos^2B = m^2.cos^2A

i.e n^2.{1−(sin^2A/n^2)} = m^2.cos^2A

[squaring (1)]

i.e n^2−sin^2A = m^2.cos^2A

i.e n^2−1+cos^2A = m^2.cos^2A

 [since sin^2A = 1 - cos^2A]

i.e. n^2−1 = m^2.cos^2A−cos^2A

i.e n^2−1 = cos2^A (m^2−1)

Hence, cos^2A = (n^2−1)/(m^2−1)

Hope it helps !!!

Answer 2

Heya friend ✌

Here is ur answer answer .....
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☢if we eliminate B.so ,we have find cosecB and cotB from the given relations and use the identity cosec^2 B-cot^2 B=1..

☣now ,tanA=ntanB....

☣cotB=n/tanA..............1)

and similarly...sinA=msinB

❣sinA=msinB.

➡cosecB=m/sinA......2)

now ,squaring 1) and 2) ,and subtracting the results we get..

m^2/sin^2 A-n^2/tan^2 A =cosec^2 B-cot^2 B

➡m^2/sin^2 A-n^2/sin^2 A/cos^2 A=1...

➡m^2/sin^2 A-n^2*cos^2 A/sin^2 A=1

➡1/sin^2 A【m^2-n^2cos^2 A)=1

➡m^2-n^2cos^2 A=1*sin^2 A

➡m^2-n^2cos^2 A=1-cos^2 A【•°•sin^2 A=1-cos^2 A】

➡-n^2Cos^2 A +cos^2 A=1-m^2【interchanging】

➡cos^2 A(-n^2+1) =(1-m^2)

➡-cos^2 A(n^2-1)=-(m^2-1)

➡cos^2 A(n^2-1)=(m^2-1)【- cancelled】

➡cos^2 A=m^2-1/n^2-1.....prooved ...

☢Hope it helps you...

@Rajukumar1!1☺