Question

# Write the value of ${\sin ^2}B + {\cos ^2}B$

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Hint: Apply Pythagoras Theorem, to find the value of hypotenuse $\left[ {{{\left( {{\text{Perpendicular}}} \right)}^2} + {{\left( {{\text{Base}}} \right)}^2} = {{\left( {{\text{Hypotenuse}}} \right)}^2}} \right]$, then use basic trigonometric ratio formula to get required answer.

In triangle ACD and in triangle BCD,$AD = 3,{\text{ }}CD = 4,{\text{ }}BD = 6$
As we see CD is perpendicular to AB, therefore triangle ACD and triangle BCD is a right angle triangle at D.
Therefore apply Pythagoras Theorem in triangle BCD
Therefore in triangle BCD
${\left( {BC} \right)^2} = {\left( {BD} \right)^2} + {\left( {DC} \right)^2} \\ \Rightarrow {\left( {BC} \right)^2} = {6^2} + {4^2} = 36 + 16 = 52 \\ \Rightarrow BC = \sqrt {52} \\$
Therefore in triangle BCD, we know$\sin B$is perpendicular divided by hypotenuse
$\Rightarrow \sin B = \dfrac{{CD}}{{BC}} \\ \Rightarrow \sin B = \dfrac{4}{{\sqrt {52} }} \\$
Now squaring both sides
$\Rightarrow {\sin ^2}B = {\left( {\dfrac{4}{{\sqrt {52} }}} \right)^2} = \dfrac{{16}}{{52}}................\left( 1 \right)$
Again in triangle BCD, we know$\cos B$is base divided by hypotenuse
$\Rightarrow \cos B = \dfrac{{BD}}{{BC}} \\ \Rightarrow \cos B = \dfrac{6}{{\sqrt {52} }} \\$
Now squaring both sides
$\Rightarrow {\cos ^2}B = {\left( {\dfrac{6}{{\sqrt {52} }}} \right)^2} = \dfrac{{36}}{{52}}................\left( 2 \right)$
Now add equation 1 and 2
$\Rightarrow {\sin ^2}B + {\cos ^2}B = \dfrac{{16}}{{52}} + \dfrac{{36}}{{52}} = \dfrac{{52}}{{52}} = 1$
So, this is the required answer.

Note: In such types of problem the key concept we have to remember is that always apply Pythagoras theorem, then calculate the value of hypotenuse using this property, then calculate the value of $\sin B$ which is perpendicular divided by hypotenuse, then calculate the value of $\cos B$ which is base divided by hypotenuse then, squaring and adding these values we will get the required answer.