Question
If $\cos\theta=\frac{12}{13},$ show that $\sin\theta(1-\tan\theta)=\frac{35}{156}.$
✓
Answer
Given: $\cos\theta=\frac{12}{13}\ \dots(1)$
To show that $\sin\theta(1-\tan\theta)=\frac{35}{156}$
Now, we know that $\cos\theta=\frac{\text{Base side adjacent to}\angle\theta}{\text{Hypotenuse}}\ \dots(2)$
Therefore, by comparing equation (1) and (2) We get, Base side adjacent to $\angle\theta=12$ And Hypotenuse = 13
Therefore from above figure Base side $BC = 12$
Hypotenuse $AC = 13$
Side AB is unknown and it can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get, $AC^2= AB^2 + BC^2$
Therefore by substituting the values of known sides
We get, $13^2 = AB^2 + 12^2$
Therefore, $\text{AB}^2=169-144$ $\text{AB}^2=25$
$\text{AB}=\sqrt{25}$ Therefore, $\text{AB}=5\ \dots(3)$
Now, we know that $\sin\theta=\frac{\text{Perpendicular side opposite to}\angle\theta}{\text{Hypotenuse}}$
Now from figure (a) We get, $\sin\theta=\frac{\text{AB}}{\text{AC}}$
Therefore, $\sin\theta=\frac{5}{13}\ \dots(4)$
Now, we know that $\tan\theta=\frac{\text{Perpendicular side opposite to}\angle\theta}{\text{Base side adjacent to }\angle\theta}$
Now from figure (a) We get, $\tan\theta=\frac{\text{AB}}{\text{BC}}$
Therefore, $\tan\theta=\frac{5}{12}\ \dots(5)$
Now L.H.S. of the equation to be proved is as follows
$\text{L.H.S.}=\sin\theta(1-\tan\theta)\ \dots(6)$
Substituting the value of $\sin\theta$ and $\tan\theta$ from equation (4) and (5) respectively We get,
$\text{L.H.S.}=\frac{5}{13}\Big(1-\frac{5}{12}\Big)$
Taking L.C.M inside the bracket We get, $\text{L.H.S.}=\frac{5}{13}\Big(\frac{1\times12}{1\times12}-\frac{5}{12}\Big)$
Therefore, $\text{L.H.S.}=\frac{5}{13}\Big(\frac{12-5}{12}\Big)$
$\text{L.H.S.}=\frac{5}{13}\Big(\frac{7}{12}\Big)$
Now, by opening the bracket and simplifying We get,
$\text{L.H.S.}=\frac{5\times7}{13\times12}$
$\text{L.H.S.}=\frac{35}{156}\ \dots(7)$ From equation (6) and (7), it can be shown that
$\sin\theta(1-\tan\theta)=\frac{35}{156}$
To show that $\sin\theta(1-\tan\theta)=\frac{35}{156}$
Now, we know that $\cos\theta=\frac{\text{Base side adjacent to}\angle\theta}{\text{Hypotenuse}}\ \dots(2)$
Therefore, by comparing equation (1) and (2) We get, Base side adjacent to $\angle\theta=12$ And Hypotenuse = 13
Therefore from above figure Base side $BC = 12$
Hypotenuse $AC = 13$
Side AB is unknown and it can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get, $AC^2= AB^2 + BC^2$
Therefore by substituting the values of known sides
We get, $13^2 = AB^2 + 12^2$
Therefore, $\text{AB}^2=169-144$ $\text{AB}^2=25$
$\text{AB}=\sqrt{25}$ Therefore, $\text{AB}=5\ \dots(3)$
Now, we know that $\sin\theta=\frac{\text{Perpendicular side opposite to}\angle\theta}{\text{Hypotenuse}}$
Now from figure (a) We get, $\sin\theta=\frac{\text{AB}}{\text{AC}}$
Therefore, $\sin\theta=\frac{5}{13}\ \dots(4)$
Now, we know that $\tan\theta=\frac{\text{Perpendicular side opposite to}\angle\theta}{\text{Base side adjacent to }\angle\theta}$
Now from figure (a) We get, $\tan\theta=\frac{\text{AB}}{\text{BC}}$
Therefore, $\tan\theta=\frac{5}{12}\ \dots(5)$
Now L.H.S. of the equation to be proved is as follows
$\text{L.H.S.}=\sin\theta(1-\tan\theta)\ \dots(6)$
Substituting the value of $\sin\theta$ and $\tan\theta$ from equation (4) and (5) respectively We get,
$\text{L.H.S.}=\frac{5}{13}\Big(1-\frac{5}{12}\Big)$
Taking L.C.M inside the bracket We get, $\text{L.H.S.}=\frac{5}{13}\Big(\frac{1\times12}{1\times12}-\frac{5}{12}\Big)$
Therefore, $\text{L.H.S.}=\frac{5}{13}\Big(\frac{12-5}{12}\Big)$
$\text{L.H.S.}=\frac{5}{13}\Big(\frac{7}{12}\Big)$
Now, by opening the bracket and simplifying We get,
$\text{L.H.S.}=\frac{5\times7}{13\times12}$
$\text{L.H.S.}=\frac{35}{156}\ \dots(7)$ From equation (6) and (7), it can be shown that
$\sin\theta(1-\tan\theta)=\frac{35}{156}$
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