Question

Plz solve question 33 and 34......correct answer will be marked as brainliest✌✌✌

Answer 1

33:
$\sin(x) + \csc(x) = 2$
squaring both sides,

$\sin^{2} (x) + 2 \sin(x) \csc(x) + \csc^{2} (x) = 4$
Since, sin(x)=1/cosec(x). Therefore sin(x)×cosec(x)=1

$\sin^{2} (x) + 2 + \csc^{2} (x) = 4 \\ \sin^{2}(x) + \csc^{2} (x) = 2$

34:

$x = b \sin(k) + a \cos(k) \\ {x}^{2} = {b}^{2} \sin^{2} (k) + 2ab \sin(k) \cos(k) + {a}^{2} \cos ^{2} (k)$
$y = a \sin(k) - b \cos(k) \\ {y}^{2} = {a}^{2} \sin^{2} (k) - 2ab \sin(k) \cos(k) + {b}^{2} \cos ^{2} (k)$
So,

$x^{2} + {y}^{2} = {b}^{2} \sin ^{2} (k) + {a}^{2} \cos^{2} (k) + {a}^{2} \sin^{2} (k) + {b}^{2} \cos^{2} (k)$
${x}^{2} + {y}^{2} = \sin ^{2} (k) ({a}^{2} + {b}^{2} ) + \cos^{2} (k)( {a}^{2} + {b}^{2} )$
${x}^{2} + {y}^{2} = ( {a}^{2} + {b}^{2} )( \sin^{2} (k) + \cos^{2} (k) )$

As
$\sin^{2} (x) + \cos^{2} (x) = 1$

Therefore,

${x}^{2} + {y}^{2} = {a}^{2} + {b}^{2}$