Question

(8)
M 2706 1माउंटेन एवरेस्ट ​

Answer 1

Given :-

• $\mathsf{ tanA = n.tanB}$

To Prove :-

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

Proof :-

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

$\qquad$❏$\;\; \mathsf{ \boxed{\mathsf{tan\theta = \dfrac{sin\theta}{cos\theta}}}}$

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

$\mathsf{We\: are \: given\pink{ : sinA = m.sinB}}$

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

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

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

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

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

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

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

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

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

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

$\qquad$❏$\;\; \mathsf{ \boxed{\mathsf{sin^2\theta + cos^2\theta = 1}}}$

$\qquad$❏$\;\; \mathsf{ \boxed{\mathsf{sin^2\theta = 1 - cos^2\theta}}}$

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

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

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

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

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