Question

If tan A =a tanB and sinA=b sinB prove that
cos²A=b^2-1/a^2-1

answer this question please​

Answer 1

Step-by-step explanation:

Given:-

• tanA = a tanB

• sinA = b sinB

To prove:-

• $\sf {cos}^{2} a = \dfrac{ {b}^{2} - 1 }{ {a}^{2} - 1}$

Proof:-

$\sf \bull \: tanA = a \: tanB$

$\sf \dashrightarrow tanB = \dfrac{1}{ a} \:tanA$

$\sf \dashrightarrow \dfrac{1}{tanB} = \dfrac{a}{ tanA}$

$\sf \dashrightarrow cotB= \dfrac{a}{ tanA}$

$\sf \bull \: sinA = b \: sinB$

$\sf \dashrightarrow sinB = \dfrac{1}{ b} \:sinA$

$\sf \dashrightarrow \dfrac{1}{sinB} = \dfrac{b}{ sinA}$

$\sf \dashrightarrow coecB= \dfrac{b}{ sinA}$

$\sf As, {cosec}^{2}B - {cot}^{2}B = 1$

$\sf \longrightarrow \bigg(\dfrac{b}{ sinA}\bigg)^{2}- \bigg(\dfrac{a}{ tanA}\bigg)^{2} = 1$

$\sf \longrightarrow \dfrac{ {b}^{2} }{ sin ^{2} A}- \dfrac{ {a}^{2} }{ tan^{2} A} = 1$

$\sf \longrightarrow \dfrac{ {b}^{2} }{ sin ^{2} A}- \dfrac{ {a}^{2} }{ \dfrac{sin^{2} A}{cos^{2} A} } = 1$

$\sf \longrightarrow \dfrac{ {b}^{2} }{ sin ^{2} A}- \dfrac{ {a}^{2} cos ^{2}A}{sin^{2}A }= 1$

$\sf \longrightarrow \dfrac{ {b}^{2} - {a}^{2} cos ^{2}A}{sin^{2}A }= 1$

$\sf \longrightarrow {b}^{2} - {a}^{2} cos ^{2}A= {sin}^{2} A$

$\sf \longrightarrow {b}^{2} - {a}^{2} cos ^{2}A= 1 - {cos}^{2} A$

$\sf \longrightarrow {b}^{2} - 1 = cos ^{2}A( {a}^{2} -1)$

$\sf \longrightarrow \dfrac{ {b}^{2} - 1}{ {a}^{2} - 1} = cos ^{2}A$

$\underline{ \boxed{\sf \leadsto cos ^{2}A = \dfrac{ {b}^{2} - 1}{ {a}^{2} - 1} }}$

$\sf \underline{Hence \: Proved}$