Question

QUESTION NUMBER : 17

Answer 1

Answer :

Given that,

a² sec²θ - b² tan²θ = c²

➨ a² sec²θ - b² (sec²θ - 1) = c²,
since sec²θ - tan²θ = 1

➨ a² sec²θ - b² sec²θ + b² = c²

➨ (a² - b²) sec²θ = c² - b²

➨ sec²θ = (c² - b²)/(a² - b²)

➨ 1/(cos²θ) = (c² - b²)/(a² - b²),
since secθ = 1/cosθ

➨ cos²θ = (a² - b²)/(c² - b²)

➨ 1 - sin²θ = (a² - b²)/(c² - b²),
since sin²θ + cos²θ = 1

➨ sin²θ = 1 - (a² - b²)/(c² - b²)

➨ sin²θ = (c² - b² - a² + b²)/(c² - b²)

➨ sin²θ = (c² - a²)/(c² - b²) [Proved]

Hope it helps!

Answer 2

$\bold {\huge{Ello!!}}$

$<b >Here's your answer$

_______________________

Given , a² sec²∅ - b² tan²∅ = c²

To prove , sin∅ = ( c² - a² ) / ( c² - b² )

Now ,

a² sec²∅ - b² tan²∅ = c²

=> a² sec²∅ - b² ( sec²∅ - 1 ) = c²

[ • We know , sec²∅ - tan²∅ = 1 → Using this identity ,
we get , tan²∅ = sec²∅ - 1 ]

=> a² sec²∅ - b² sec²∅ + b² = c²

=> sec²∅ ( a² - b² ) = ( c² - b² )

=> sec²∅ = ( c² - b² ) / ( a² - b² )

[ • We know , sec∅ = 1 / cos∅ . so , sec²∅ = 1 / cos²∅ ]

=> 1 / cos²∅ = ( c² - b² ) / ( a² - b² )

=> cos²∅ = ( a² - b² ) / ( c² - b² )

[ • sin²∅ + cos²∅ = 1 → Using this identity ,
we get , cos²∅ = 1 - sin²∅ ]

=> 1 - sin²∅ = ( a² - b² ) / ( c² - b² )

=> sin²∅ = 1 - { ( a² - b² ) / ( c² - b² ) }

=> sin²∅ = { ( c² - b² ) - ( a² - b² ) } / ( c² - b² )

=> sin²∅ = ( c² - b² - a² + b² ) / ( c² - b² )

=> sin²∅ = ( c² - a² ) / ( c² - b² ) [ ★ Hence Proved ]


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$<b><u><marquee direction> Hope it helps !!$