If cos alpha/COS beta=m and cos …

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Sia ? 6 years, 5 months ago
Given, m = {tex}\frac { \cos \alpha } { \cos \beta }{/tex} ......(1) and, n = {tex}\frac { \cos \alpha } { \sin \beta }{/tex}.......(2)
LHS = (m2 +n2) cos2{tex}\beta{/tex}
{tex}\Rightarrow \mathrm { LHS } = \left( \frac { \cos ^ { 2 } \alpha } { \cos ^ { 2 } \beta } + \frac { \cos ^ { 2 } \alpha } { \sin ^ { 2 } \beta } \right) \cos ^ { 2 } \beta{/tex} [ from (1) & (2) ]
{tex}\Rightarrow \mathrm { LHS } = \left( \frac { \cos ^ { 2 } \alpha \sin ^ { 2 } \beta + \cos ^ { 2 } \alpha \cos ^ { 2 } \beta } { \cos ^ { 2 } \beta \sin ^ { 2 } \beta } \right) \cos ^ { 2 } \beta{/tex}
{tex}\Rightarrow \mathrm { LHS } = \cos ^ { 2 } \alpha \left( \frac { \sin ^ { 2 } \beta + \cos ^ { 2 } \beta } { \cos ^ { 2 } \beta \sin ^ { 2 } \beta } \right) \cos ^ { 2 } \beta{/tex}
{tex}\Rightarrow \mathrm { LHS } = \cos ^ { 2 } \alpha \left( \frac { 1 } { \cos ^ { 2 } \beta \sin ^ { 2 } \beta } \right) \cos ^ { 2 } \beta{/tex} [ Since, sin2A + cos2A =1]
{tex}\Rightarrow L.H.S. = \frac { \cos ^ { 2 } \alpha } { \sin ^ { 2 } \beta } = \left( \frac { \cos \alpha } { \sin \beta } \right) ^ { 2 }{/tex}
{tex}\therefore L.H.S. = n^2{/tex} [ from (2) ]
= R.H.S . Hence, Proved.
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