Question

$\huge\red{\boxed{\sf QuEStiON }}$

If Tan(A)=n Tan(B) and Sin(A)=m Sin(B)
Then,
Prove that
Cos²(A)=m²-1 n²-1​

Answer 1

Answer:

$\mathcal{ANSWER}$

We have,

tan α = n tan β

⇒ tan β = tan α/n

⇒ cot β = n/tan α

sin α = m sin β

⇒ sin β = sin α/m

⇒ cosec β = m/sin α

Since ,

cosec2 β – cot2 β = 1

⇒ m2/sin2 α – n2/tan2 α = 1

⇒ m2/sin2 α – n2 cos2 α/sin2 α = 1

⇒ m2 – n2 cos2 α = sin2 α

⇒ m2 – n2 cos2 α = 1 – cos2 α

⇒ m2 – 1 = (n2 – 1) cos2 α

⇒ cos2 α = (m2 – 1)/(n2 – 1)

Hence proved.

Answer 2

Answer:

In this question we have to find cos A in terms of m and n, so we have to eliminate ZB from the given relations.

tan A = n tan B tan B = 1/n tan A

Cot B = n /tan A [ cot B = 1/tan B]

sin A = m sinB sin B = 1/m sinA

cosec B = m/ sinA

cosec?A - cot B =1

Substitute the value of cot B and cosec B in the above relation.

(m/ sinA) - (n /tan A)? (m2 / sin A) - (n? /tan? A) (m? / sin?A) - (n? /(sin?A / cos A))

[ tan A = sinA / cosA]

(m? / sin?A) - n°cos?A / sin?A = 1 m2 - n?cos?A = sin?A =

m? - n?cos?A = 1- cos a

(m² / sin?A) - (n? /(sin?A / cos A))

[ tan A = sinA / cosA]

(m2 / sin?A) - n°cos A/ sin?A = 1 m? - n°cosA = sin?A =

m? - n?cos?A = 1- cos?A

[sin°A = 1- cos A]

m2 -1 = n°cos A - cos?A m2 - 1= cos 2A(n2 -1)

cos?A = m2 -1/ n²-1

HOPE THIS WILL HELP YOU...