Question

77.) If osec 0 - sineO = a^3, secO - cosO = b prove that a^2 b^2(a^2 + b^2) = 1​

Answer 1

Heya friend,

Given :-

cosec ∅ - sin ∅ = $a^{3}$

sec ∅ - cos ∅ = $b^{3}$

To prove :- $a^{2} b^{2} (a^{2} +b^{2}) = 1$

Now,

(1/sin ∅) - sin ∅ = $a^{3}$

(1-$sin^{2}$∅) / sin∅ = $a^{3}$

$cos^{2}$∅ / sin∅ = $a^{3}$

Now, we know from the question that we require $a^{2}$, so first we will cube root $a^{3}$ to get a and then square it .

So,

$(cos^{2})^{\frac{1}{3} }$ ∅ / $sin^{\frac{1}{3} }$∅ = $a^{3}$

Now, squaring both sides, we get,

$cos^{\frac{4}{3} }$ ∅ / $sin^{\frac{2}{3} }$ ∅ = $a^{2}$ -------------------- (1)

and,

sec ∅ - cos ∅ = $b^{3}$

(1 / cos∅) - cos ∅ = $b^{3}$

( 1 - $cos^{2}$ ∅ ) / cos ∅ = $b^{3}$

$sin^{2}$ ∅ / cos ∅ = $b^{3}$

Again,

$sin^{\frac{2}{3} }$ ∅ / $cos^{\frac{1}{3} }$ = b

Squaring both sides,

$sin^{\frac{4}{3} }$ ∅  / $cos^{\frac{2}{3} }$ ∅ = $b^{2}$ ---------------------- (2)

Now from L.H.S. ,

$a^{2} b^{2}$ =  ( $cos^{\frac{4}{3} }$ ∅ / $sin^{\frac{2}{3} }$ ∅ ) × ( $sin^{\frac{4}{3} }$ / $cos^{\frac{2}{3} }$ )

      = $cos^{\frac{2}{3} }$ ∅ $sin^{\frac{2}{3} }$ ∅

and,

$a^{2} + b^{2}$ = ( $cos^{\frac{4}{3} }$ ∅ / $sin^{\frac{2}{3} }$ ∅ ) + ( $sin^{\frac{4}{3} }$ ∅ / $cos^{\frac{2}{3} }$ ∅ )

           = [ $cos^{\frac{4}{3} + \frac{2}{3} }$ + $sin^{\frac{2}{3} +\frac{4}{3} }$ ] / $sin^{\frac{2}{3} }$∅ $cos^{\frac{2}{3} }$ ∅

           = ( $sin^{2}$∅ +$cos^{2}$∅ ) / $sin^{\frac{2}{3} }$ ∅ $cos^{\frac{2}{3} }$ ∅

           = 1 / $sin^{\frac{2}{3} }$ ∅  $cos^{\frac{2}{3} }$∅

Now,

$a^{2} b^{2} (a^{2} +b^{2})$ =

⇒ $sin^{\frac{2}{3} }$ ∅ $cos^{\frac{2}{3} }$ ∅ × ( 1 / $sin^{\frac{2}{3} }$ ∅ $cos^{\frac{2}{3} }$ ∅ )

⇒ 1

Hence proved !!!

Thanks !

#BAL #answerwithquality