From the given figure, Find ${cos}^{2} A + {sin}^{2} A$

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $1$
Given, $△ A B C$ , $∠ A B C = 90^{\circ}$ , $A B = B C = a$
By Pythagoras Theorem,
$A C^{2} = A B^{2} + B C^{2}$
$A C^{2} = a^{2} + a^{2}$
$A C = a \sqrt{2}$
Now, ${cos}^{2} A + {sin}^{2} A = (\frac{B}{H})^{2} + (\frac{P}{H})^{2} = (\frac{a}{a \sqrt{2}})^{2} + (\frac{a}{a \sqrt{2}})^{2}$
= $\frac{1}{2} + \frac{1}{2}$
= $1$

Suggest Corrections

Similar questions

{cos}^{2} A + {sin}^{2} A

In the given figure, $\frac{A B}{B C}$ = $\frac{3}{4}$ .Find the value of $s i n^{2} A$ + $c o s^{2} A$ =

\frac{{sin}^{2} A}{{cos}^{2} A} + \frac{{cos}^{2} a}{{sin}^{2} A} = \frac{1}{{sin}^{2} A {cos}^{2} A} - 2

How many of below given statements are correct?

1. sin2A = 2sinA.cosA

2.cos2A = $s i n^{2} A - c o s^{2} A$

3.tan A = $\frac{2 t a n \frac{A}{2}}{1 - t a n^{2} \frac{A}{2}}$

4. $s i n^{2} \frac{A}{2}$ = 1 - cosA

5. $t a n^{2} A$ = $\frac{1 - c o s 2 A}{1 + c o s 2 A}$

6. $s i n^{2} A$ = $\frac{2 t a n A}{1 - t a n^{2} A}$

\frac{s i n^{2} A}{c o s^{2} A} + \frac{c o s^{2} A}{s i n^{2} A} = \frac{1}{s i n^{2} A c o s^{2} A} - 2

Join BYJU'S Learning Program