Question
If $\cot \theta=1$; find the value of: $5 \tan ^2 \theta+2 \sin ^2 \theta-3$
✓
Answer
Consider the diagram below :
$\cot θ = 1$
i.e. $\frac{\text { base }}{\text { perpendicular }}=\frac{1}{1}$
Therefore if length of base $= x,$ length of perpendicular $= x$
Since
base$^2 +$ perpendicular$^2 =$ hypotenuse$^2 \dots...[$ Using Pythagooras Theorem$]$
$(x)^2 + (x)^2 =$ hypotenuse$^2$
hypotenuse$^2 = x^2 + x^2 = 2x$
hypotenuse $=\sqrt{2} x$
Now
$\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{x}{\sqrt{2} x}=\frac{1}{\sqrt{2}}$
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{x}{x}=1$
Therefore
$5 \tan^2 \theta + 2\sin^2 \theta – 3$
$=5(1)^2+2\left(\frac{1}{\sqrt{2}}\right)^2-3$
$=5+1-3$
$=3$
$\cot θ = 1$
i.e. $\frac{\text { base }}{\text { perpendicular }}=\frac{1}{1}$
Therefore if length of base $= x,$ length of perpendicular $= x$
Since
base$^2 +$ perpendicular$^2 =$ hypotenuse$^2 \dots...[$ Using Pythagooras Theorem$]$
$(x)^2 + (x)^2 =$ hypotenuse$^2$
hypotenuse$^2 = x^2 + x^2 = 2x$
hypotenuse $=\sqrt{2} x$
Now
$\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{x}{\sqrt{2} x}=\frac{1}{\sqrt{2}}$
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{x}{x}=1$
Therefore
$5 \tan^2 \theta + 2\sin^2 \theta – 3$
$=5(1)^2+2\left(\frac{1}{\sqrt{2}}\right)^2-3$
$=5+1-3$
$=3$
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