Question

If Sin * theta = X/Y then Cos '0' is
A. y/(sqrt(y ^ 2 - x ^ 2));
B. y/x;
C. x/(sqrt(y ^ 2 - x ^ 2))
D. (sqrt(y ^ 2 - x ^ 2))/y
plz answer me correctly yaar​

Answer 1

Answer:

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Answer 2

$sinθ =\frac{sideoppositetoangleθ​ }{hypotenuse}$

or sinθ= $\frac{Perpendicular​ }{hypotenuse}$

sin θ =$\frac{x^{2} - y^{2} }{x^{2} + y^{2} }$  ⇒$\frac{P}{H}$ = $\frac{x^{2}- y^{2} }{x^{2} + y^{2} }$  

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⇒  $\frac{AB}{BC}$ = ​$\frac{x^{2} -y^{2} }{x^{2} +y^{2} }$    

side opposite to angle θ=AB=$x^{2} - y^{2}$

hypotenuse AC=$x^{2} + y^{2}$

In right angled △ABC, we have

⇒$(AB)^{2}$ + $(BC)^{2}$=$(AC)^{2}$⇒ $(x^{2}+y^{2} )$+

⇒$(x^{2} +y^{2}) ^{2}$   ⇒$(BC)^{2}$=$(x^{2}+ y^{2} )^{2}$−$(x^{2} -y^{2} )^{2}$(By Pythagoras theorem)

⇒$(BC)^{2}$ =  $[x^{2}+y^{2} + x^{2}- y^{2}]$  $[x^{2} +y^{2} (x^{2}- y^{2} )]$  = $(2x ^{2} )$  $(2y^{2} )$

(using identity $a^{2}$−$b^{2}$ =(a+b)(a−b))

BC=$\sqrt{4x} ^{2} y^{2}$ = ±2xy

taking positie square root since, side cannot be negative

cosθ=  $\frac{Base}{Hypotenuse}$  =$\frac{BC}{AC}$  = $\frac{2xy}{x^{2} +y^{2} }$

and tanθ=  $\frac{Perpendicular}{Base}$  =  $\frac{AB}{BC}$  = $\frac{x^{2} - y^{2} }{2xy}$

so,$\frac{1}{tanθ}$  =   $\frac{1}{x^{2}- y^{2} }$  =  $\frac{2xy}{x^{2} - y^{2} }$

                   $\frac{}{2xy}$