Question

ques 5th plz solve...............

Answer 1

Explanation:

Given that, tanA=ntanB⇒sinAcosA=n⋅sinBcosB.

⇒sinAsinB=n⋅cosAcosB.....................................⟨1⟩.

Also given that, sinA=msinB⇒sinAsinB=m.....................⟨2⟩.

Comparing ⟨1⟩and⟨2⟩, we get,

m=n⋅cosAcosB, giving, cosB=nm⋅cosA......⟨3⟩.

⟨2⟩⇒sinB=1m⋅sinA...................................⟨2'⟩.

Now, using ⟨2'⟩and⟨3⟩ in cos2B+sin2B=1, we get,

⇒n2m2⋅cos2A+1m2⋅sin2A=1.

⇒n2cos2A+sin2A=m2.

⇒n2cos2A+(1−cos2A)=m2.

⇒n2cos2A−cos2A=m2−1,i.e.,

⇒(n2−1)cos2A=(m2−1).

⇒cos2A=m2−1n2−1, as desired!

Answer 2

$\mathsf{Given : tanA = n.tanB}$

$\bigstar\;\; \mathsf{We\;know\;that : \boxed{\mathsf{tan\theta = \dfrac{sin\theta}{cos\theta}}}}$

$\implies \sf{\dfrac{sinA}{cosA} = (n)\bigg(\dfrac{sinB}{cosB}\bigg)}$

$\mathsf{Given : sinA = m.sinB}$

$\implies \sf{(m)\bigg(\dfrac{sinB}{cosA}\bigg) = (n)\bigg(\dfrac{sinB}{cosB}\bigg)}$

$\implies \sf{\bigg(\dfrac{m}{cosA}\bigg) = \bigg(\dfrac{n}{cosB}\bigg)}$

$\implies \sf{cosB = \bigg(\dfrac{n}{m}\bigg)(cosA)}$

$\textsf{Squaring on both sides, We get :}$

$\implies \sf{cos^2B = \bigg(\dfrac{n^2}{m^2}\bigg)(cos^2A)\;------\;[1]}$

$\textsf{Now, Consider : sinA = m.sinB}$

$\implies \mathsf{sinB = \dfrac{sinA}{m}}$

$\textsf{Squaring on both sides, We get :}$

$\implies \mathsf{sin^2B = \dfrac{sin^2A}{m^2}\;------\;[2]}$

$\textsf{Adding Both Equations [1] and [2], We get :}$

$\implies \mathsf{sin^2B + cos^2B = \bigg(\dfrac{sin^2A}{m^2}\bigg) + \bigg(\dfrac{n^2}{m^2}\bigg)(cos^2A)}$

$\bigstar\;\; \mathsf{We\;know\;that : \boxed{\mathsf{sin^2\theta + cos^2\theta = 1}}}$

$\bigstar\;\; \mathsf{We\;know\;that : \boxed{\mathsf{sin^2\theta = 1 - cos^2\theta}}}$

$\implies \mathsf{\bigg(\dfrac{1 - cos^2A}{m^2}\bigg) + \bigg(\dfrac{n^2}{m^2}\bigg)(cos^2A) = 1}$

$\implies \mathsf{({1 - cos^2A}) + (n^2)(cos^2A) = m^2}$

$\implies \mathsf{(n^2)(cos^2A) - (cos^2A) = (m^2 - 1)}$

$\implies \mathsf{(cos^2A)(n^2 - 1) = (m^2 - 1)}$

$\implies \mathsf{cos^2A = \dfrac{m^2 - 1}{n^2 - 1}}$