Question

Plz ans the que in attachment

Class X, Introduction to Trigonometry

Answer 1

$\textsf{\underline {\Large {TRIGONOMETRY}}}$ :

( iii ). For $\mathsf{\dfrac{x} {a} cosA \:+\:{\dfrac{y} {b} sinA\:=1}}$

Squaring both sides,

$\mathsf[{\dfrac{x}{a} cosA}\:+\:{\mathsf{\dfrac{y} {b} sinA]} ^{2} \:=\:{1}^{2}}$

$\mathsf{\dfrac{{x}^{2}} {{a}^{2}} {cos}^{2}A\:+\:{\dfrac{{y} ^{2}} {{b}^{2}} {sin}^{2}A\:+\:{\dfrac{2xy \:cos A sinA} {ab} \:=1}}}$ --> ( i )

Similarly, For $\mathsf{\dfrac{x} {a} sinA\:-\:{\dfrac{y} {b} cosA\:=\:=1}}$

Squaring both sides,

$\mathsf[{\dfrac{x}{a} cosA}\:-\:{\mathsf{\dfrac{y} {b} sinA]} ^{2} \:=\:{1}^{2}}$

$\mathsf{\dfrac{{x}^{2} }{{a}^{2}} {sin}^{2}A\:+\:{\dfrac{{y}^{2}}{{b}^{2}} {cos}^{2}A\:-\:{\dfrac{2xy \:cos A sinA} {ab} \:=1}}}$ --> ( ii )

Now, Adding ( i ) and ( ii ),

$\mathsf{\dfrac{{x}^{2}}{{a}^{2}} ({sin}^{2}A\:+\:{cos} ^{2}A) \:+\:{\dfrac{{y}^{2}}{{b}^{2}}( {cos}^{2}A\:+\:{sin} ^{2}A)\:= \:1\:+1\:}}$

We know that,

$\boxed{\mathsf{{sin} ^{2}A \:+\:{cos} ^{2}A\:=\:1}}$

➡️ $\boxed{ \mathsf{\dfrac{{x}^{2}}{{a}^{2}} \:+\:{\dfrac{{y}^{2}}{{b}^{2}}\:=\:2}}}$

$\tt{\large{Hence, Proved.}}$