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If tera is acute and (1-a^2)sin teta=(1+a^2)cos teta then sin teta=

Ajay
209 Points
4 years ago
Solved below.........................................................................................................
$Given\quad (1-{ a }^{ 2 })sin\theta \quad =\quad (1+{ a }^{ 2 })cos\theta \\ squaring\quad both\quad sides\\ { (1-{ a }^{ 2 }) }^{ 2 }{ sin }^{ 2 }\theta \quad =\quad { (1+{ a }^{ 2 }) }^{ 2 }{ cos }^{ 2 }\theta \quad =\quad { (1+{ a }^{ 2 }) }^{ 2 }{ (1-sin }^{ 2 }\theta )\\ \left\{ { (1-{ a }^{ 2 }) }^{ 2 }+{ (1+{ a }^{ 2 }) }^{ 2 } \right\} { sin }^{ 2 }\theta \quad =\quad { (1+{ a }^{ 2 }) }^{ 2 }\\ { sin }^{ 2 }\theta \quad =\quad \frac { { (1+{ a }^{ 2 }) } }{ 2(1+{ a }^{ 4 }) } \\ sin\theta \quad =\quad \begin{matrix} + \\ - \end{matrix}\sqrt { \frac { { (1+{ a }^{ 2 }) } }{ 2(1+{ a }^{ 4 }) } } \\ since\quad \theta \quad is\quad acute\quad ignoring\quad the\quad negative\quad sign\quad hence\\ sin\theta \quad =\quad \sqrt { \frac { { (1+{ a }^{ 2 }) } }{ 2(1+{ a }^{ 4 }) } }$
Ajay
209 Points
4 years ago
There was mistake in previous solution, posting again......................................................
$Given\quad (1-{ a }^{ 2 })sin\theta \quad =\quad (1+{ a }^{ 2 })cos\theta \\ squaring\quad both\quad sides\\ { (1-{ a }^{ 2 }) }^{ 2 }{ sin }^{ 2 }\theta \quad =\quad { (1+{ a }^{ 2 }) }^{ 2 }{ cos }^{ 2 }\theta \quad =\quad { (1+{ a }^{ 2 }) }^{ 2 }{ (1-sin }^{ 2 }\theta )\\ \left\{ { (1-{ a }^{ 2 }) }^{ 2 }+{ (1+{ a }^{ 2 }) }^{ 2 } \right\} { sin }^{ 2 }\theta \quad =\quad { (1+{ a }^{ 2 }) }^{ 2 }\\ { sin }^{ 2 }\theta \quad =\quad \frac { { { (1+{ a }^{ 2 }) }^{ 2 } } }{ 2(1+{ a }^{ 4 }) } \\ sin\theta \quad =\quad \begin{matrix} + \\ - \end{matrix}\quad \frac { (1+{ a }^{ 2 }) }{ \sqrt { 2(1+{ a }^{ 4 }) } } \\ since\quad \theta \quad is\quad acute\quad ignoring\quad the\quad negative\quad sign\quad hence\\ sin\theta \quad =\quad \quad \frac { (1+{ a }^{ 2 }) }{ \sqrt { 2(1+{ a }^{ 4 }) } }$