Question

if cotA=20/21. find sin^2A+cos^2A​

Answer 1

S O L U T I O N :

Firstly, attachment a figure of right angled triangle according to the question.

$\underline{\bf{Given\::}}$

Cot A = 20/21

$\underline{\bf{Explanation\::}}$

As we know that cot Ф;

$\boxed{\bf{cot\:\theta = \frac{Base}{Perpendicular} }}$

$\mapsto\tt{cot\:A = \dfrac{BC}{AB}=\dfrac{20}{21} }$

$\underline{\underline{\tt{Using\:by\:Pythagoras\:theorem\::}}}$

$\longrightarrow\sf{(hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2}}$

$\longrightarrow\sf{(AC)^{2} = (BC)^{2} + (AB)^{2}}$

$\longrightarrow\sf{(AC)^{2} = (20)^{2} + (21)^{2}}$

$\longrightarrow\sf{(AC)^{2} = 400 + 441}$

$\longrightarrow\sf{(AC)^{2} = 841}$

$\longrightarrow\sf{AC=\sqrt{841} }$

$\longrightarrow\bf{AC= 29\:cm}$

Now,

$\mapsto\tt{sin^{2} \:A + cos^{2} \:A}$

$\mapsto\tt{\bigg(\dfrac{Perpendicular}{Hypotenuse } \bigg)^{2} + \bigg(\dfrac{Base}{Hypotenuse }\bigg)^{2}}$

$\mapsto\tt{\bigg(\dfrac{AB}{AC} \bigg)^{2} + \bigg(\dfrac{BC}{AC} \bigg)^{2}}$

$\mapsto\tt{\bigg(\dfrac{21}{29} \bigg)^{2} + \bigg(\dfrac{20}{29} \bigg)^{2}}$

$\mapsto\tt{\dfrac{441}{841}+\dfrac{400}{841}}$

$\mapsto\tt{\dfrac{441+ 400}{841}}$

$\mapsto\tt{\cancel{\dfrac{841}{841}}}$

$\mapsto\bf{1}$

Thus,

The value of sin²A + cos²A  will be 1 .