Question

If tanA=n tanB and sinA=m sinB,prove that cos²A=m²-1/n²-1 [Don't spam]

Answer 1

Answer:

I think the queues should be

Prove  2=(2−1)/(2−1)cos2A=(n2−1)/(m2−1)

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:

2.2=2.2n2.cos2B=m2.cos2A

i.e  2.(1−2)=2.2n2.(1−sin2B)=m2.cos2A

i.e  2.(1−(2/2))=2.2n2.(1−(sin2A/n2))=m2.cos2A  …….[squaring (1)]

i.e  2−2=2.2n2−sin2A=m2.cos2A

i.e  2−1+2=2.2n2−1+cos2A=m2.cos2A  [since sin^2A = 1 - cos^2A]

i.e  2−1=2.2−2n2−1=m2.cos2A−cos2A

i.e  2−1=2(2−1)n2−1=cos2A(m2−1)

Hence,  2=(2−1)/(2−1)

Step-by-step explanation:

Answer 2

☆ANSWER ☆

Hint - We have to find cos2 A in terms of m and  n. This means that the angle B is to be eliminated from the given relations.

☆SOLUTION: - REFER TO THE ATTACHMENT

HOPE IT HELPS :)