Question

Prove that sin? + cos²0 =1.​

Answer 1

Answer:

your answer it will help u

Answer 2

Appropriate Question -:

Prove that:

$\sf{{sin}^{2} \theta + {cos}^{2} \theta = 1}$

Answer :-

Let's take a right-angled triangle ABC, where angle B = 90°,

FIGURE :-

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf C}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf A}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf \Theta }\end{picture}$

• Refer the attachment.

We know,

$\sf{sine(sin) = \dfrac{Perpendicular}{Hypotenuse}}$

$\sf{cosine(cos) = \dfrac{base}{Hypotenuse}}$

Therefore,

$\longrightarrow \: \: \sf{sin \theta = \dfrac{BC}{AC}}$

And

$\longrightarrow \: \: \sf{cos \theta = \dfrac{AB}{AC}}$

Now,

(sin²θ + cos²θ) = $\sf{\bigg( \dfrac{BC}{AC} \bigg)}$ + $\sf{\bigg( \dfrac{AB}{AC} \bigg)}$

$\sf{ \longrightarrow \: {\bigg( \dfrac{BC}{AC} \bigg)}^{2} + {\bigg( \dfrac{AB}{AC} \bigg)}^{2}}$

$\sf{ \longrightarrow \: \dfrac{{BC}^{2} + {AB}^{2}}{{AC}^{2}}}$

$\sf{ \longrightarrow \: \dfrac{{AC}^{2}}{{AC}^{2}}}$

[AC² in numerator ∵ BC² + AB² = AC²],

$\sf{ \longrightarrow \: \cancel{ \dfrac{{AC}^{2}}{{AC}^{2}}}}$

$\sf{ \longrightarrow \: 1}$

Hence,

$\dag \: \: \: \: \: \: \: \bf{{sin}^{2} \theta + {cos}^{2} \theta = 1}$