
Prove that \[2{\sin ^2}\theta + 3{\cos ^2}\theta = 2 + {\cos ^2}\theta \]
Answer
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Hint: In this question, we have to prove that the given relation is equal or not. The given problem is the relation to prove. By using the trigonometry relations in the given relation we will prove the required result. We have to apply the formula of ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Complete step-by-step answer:
It is stated in the question \[2{\sin ^2}\theta + 3{\cos ^2}\theta \]….$(i)$
Now we have to apply the formula of ${\sin ^2}\theta + {\cos ^2}\theta = 1$ from we here we can write ${\sin ^2}\theta = 1 - {\cos ^2}\theta $ which we have to put in the above equation $(i)$
\[2{\sin ^2}\theta + 3{\cos ^2}\theta \]
By using the relation ${\sin ^2}\theta + {\cos ^2}\theta = 1$ we get,
$ = 2(1 - {\cos ^2}\theta ) + 3{\cos ^2}\theta $
Multiplying the terms we get,
$ = 2 - 2{\cos ^2}\theta + 3{\cos ^2}\theta $
Simplifying we get,
$ = 2 + {\cos ^2}\theta $
$\therefore $ We have proved \[2{\sin ^2}\theta + 3{\cos ^2}\theta = 2 + {\cos ^2}\theta \].
Note: For solving this questions of trigonometry you have to remember the formulas for all like if you have to find out the value of $\sin \theta $ or $\operatorname{Cos} \theta $ you can apply the formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$ if you have the value of either of them
Next if you need to find out the value of $\sec \theta $ or $\tan \theta $ then you can apply the formula ${\sec ^2}\theta - {\tan ^2}\theta = 1$, if you have the value of either of them
Now if you want to find out the value of $\cot \theta $ or $\cos ec\theta $, then you can apply the formula $\cos e{c^2}\theta - {\cot ^2}\theta = 1$, but for this you must know the value of either of them.
Besides these there are some other formulas which are necessary for solving the questions of trigonometry like $\sin \theta = \dfrac{1}{{\cos ec\theta }}$, $\tan \theta = \dfrac{1}{{\cot \theta }}$ and $\cos \theta = \dfrac{1}{{\sec \theta }}$
In some cases there are some questions where we have to find out the value of $\theta $ by applying those formula like $\sin \theta = \cos ({90^ \circ } - \theta )$, $\tan \theta = \cot ({90^ \circ } - \theta )$ sand $\sec \theta = \cos ec({90^ \circ } - \theta )$
Complete step-by-step answer:
It is stated in the question \[2{\sin ^2}\theta + 3{\cos ^2}\theta \]….$(i)$
Now we have to apply the formula of ${\sin ^2}\theta + {\cos ^2}\theta = 1$ from we here we can write ${\sin ^2}\theta = 1 - {\cos ^2}\theta $ which we have to put in the above equation $(i)$
\[2{\sin ^2}\theta + 3{\cos ^2}\theta \]
By using the relation ${\sin ^2}\theta + {\cos ^2}\theta = 1$ we get,
$ = 2(1 - {\cos ^2}\theta ) + 3{\cos ^2}\theta $
Multiplying the terms we get,
$ = 2 - 2{\cos ^2}\theta + 3{\cos ^2}\theta $
Simplifying we get,
$ = 2 + {\cos ^2}\theta $
$\therefore $ We have proved \[2{\sin ^2}\theta + 3{\cos ^2}\theta = 2 + {\cos ^2}\theta \].
Note: For solving this questions of trigonometry you have to remember the formulas for all like if you have to find out the value of $\sin \theta $ or $\operatorname{Cos} \theta $ you can apply the formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$ if you have the value of either of them
Next if you need to find out the value of $\sec \theta $ or $\tan \theta $ then you can apply the formula ${\sec ^2}\theta - {\tan ^2}\theta = 1$, if you have the value of either of them
Now if you want to find out the value of $\cot \theta $ or $\cos ec\theta $, then you can apply the formula $\cos e{c^2}\theta - {\cot ^2}\theta = 1$, but for this you must know the value of either of them.
Besides these there are some other formulas which are necessary for solving the questions of trigonometry like $\sin \theta = \dfrac{1}{{\cos ec\theta }}$, $\tan \theta = \dfrac{1}{{\cot \theta }}$ and $\cos \theta = \dfrac{1}{{\sec \theta }}$
In some cases there are some questions where we have to find out the value of $\theta $ by applying those formula like $\sin \theta = \cos ({90^ \circ } - \theta )$, $\tan \theta = \cot ({90^ \circ } - \theta )$ sand $\sec \theta = \cos ec({90^ \circ } - \theta )$
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