Answer
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Hint: Some basic identities to be used in this question are as follows:-
1. $\sec \theta = \dfrac{1}{{\cos \theta }}\\$
2. $\cos ec\theta = \dfrac{1}{{\sin \theta }}\\$
3. ${\sec ^2}\theta = 1 + {\tan ^2}\theta \\$
4. $\cos e{c^2}\theta = 1 + {\cot ^2}\theta $
Complete step-by-step answer:
$2{\sec ^2}\theta - {\sec ^4}\theta - 2\cos e{c^2}\theta + \cos e{c^4}\theta = {\cot ^4}\theta - {\tan ^4}\theta $
For this question we will take L.H.S part and convert it in the form of R.H.S for this Firstly we will separate the same degree terms with each other.
\[ = 2({\sec ^2}\theta - \cos e{c^2}\theta ) - {\sec ^4}\theta + \cos e{c^4}\theta \] (Taking 2 as common)
\[ = 2[1 + {\tan ^2}\theta - (1 + {\cot ^2}\theta )] - {\sec ^4}\theta + \cos e{c^4}\theta \] (Putting identity number 3 and 4 mentioned in above hint)
\[ = 2[1 + {\tan ^2}\theta - 1 - {\cot ^2}\theta ] - {\sec ^4}\theta + \cos e{c^4}\theta \]
Multiplying 2 with the rest of the bracket terms.
\[ = 2 + 2{\tan ^2}\theta - 2 - 2{\cot ^2}\theta - [{({\sec ^2}\theta )^2}] + [{(\cos e{c^2}\theta )^2}]\]
Splitting degree of \[{\sec ^4}\theta \] as \[{({\sec ^2}\theta )^2}\]and \[\cos e{c^4}\theta \] as \[{(\cos e{c^2}\theta )^2}\] so that we can use identities there. If not done then it would be difficult to solve the quadruple degree.
\[ = 2{\tan ^2}\theta - 2{\cot ^2}\theta - [{(1 + {\tan ^2}\theta )^2}] + [{(1 + {\cot ^2}\theta )^2}]\] (Again using identities mentioned in Hint number 3 and 4)
\[ = 2{\tan ^2}\theta - 2{\cot ^2}\theta - [1 + {\tan ^4}\theta + 2{\tan ^2}\theta ] + [1 + {\cot ^4}\theta + 2{\cot ^2}\theta ]\]
Using identity ${(a + b)^2} = {a^2} + {b^2} + 2ab$
\[ = 2{\tan ^2}\theta - 2{\cot ^2}\theta - 1 - {\tan ^4}\theta - 2{\tan ^2}\theta + 1 + {\cot ^4}\theta + 2{\cot ^2}\theta \]
Cancelling alike and opposite terms we will get the required answer
\[ = {\cot ^4}\theta - {\tan ^4}\theta \]
Hence proved.
Note:
1. This question can be done by various methods one method is by converting the whole equation into $\sin \theta $ and $\cos \theta $. But it will be very long and time consuming. The approach by which we have solved is way more convenient rather than converting the whole into $\sin \theta $ and $\cos \theta $. So some identities need to be memorised so that it can be done easily and fast.
2. Be very cautious when opening the degree of a trigonometric function. For example: - ${\sec ^4}\theta = {\left( {{{\sec }^2}\theta } \right)^2}$
Importance of trigonometric functions: - In geometry trigonometric functions are used to find unknown angles or sides of right angled triangles. The three common used trigonometric functions are $\sin \theta ,\cos \theta ,\tan \theta $.If we have good understanding of these three functions then other trigonometric functions such as $\cos ec\theta ,\sec \theta ,\cot \theta $ which are reciprocal of the above three functions can b easily understood.
1. $\sec \theta = \dfrac{1}{{\cos \theta }}\\$
2. $\cos ec\theta = \dfrac{1}{{\sin \theta }}\\$
3. ${\sec ^2}\theta = 1 + {\tan ^2}\theta \\$
4. $\cos e{c^2}\theta = 1 + {\cot ^2}\theta $
Complete step-by-step answer:
$2{\sec ^2}\theta - {\sec ^4}\theta - 2\cos e{c^2}\theta + \cos e{c^4}\theta = {\cot ^4}\theta - {\tan ^4}\theta $
For this question we will take L.H.S part and convert it in the form of R.H.S for this Firstly we will separate the same degree terms with each other.
\[ = 2({\sec ^2}\theta - \cos e{c^2}\theta ) - {\sec ^4}\theta + \cos e{c^4}\theta \] (Taking 2 as common)
\[ = 2[1 + {\tan ^2}\theta - (1 + {\cot ^2}\theta )] - {\sec ^4}\theta + \cos e{c^4}\theta \] (Putting identity number 3 and 4 mentioned in above hint)
\[ = 2[1 + {\tan ^2}\theta - 1 - {\cot ^2}\theta ] - {\sec ^4}\theta + \cos e{c^4}\theta \]
Multiplying 2 with the rest of the bracket terms.
\[ = 2 + 2{\tan ^2}\theta - 2 - 2{\cot ^2}\theta - [{({\sec ^2}\theta )^2}] + [{(\cos e{c^2}\theta )^2}]\]
Splitting degree of \[{\sec ^4}\theta \] as \[{({\sec ^2}\theta )^2}\]and \[\cos e{c^4}\theta \] as \[{(\cos e{c^2}\theta )^2}\] so that we can use identities there. If not done then it would be difficult to solve the quadruple degree.
\[ = 2{\tan ^2}\theta - 2{\cot ^2}\theta - [{(1 + {\tan ^2}\theta )^2}] + [{(1 + {\cot ^2}\theta )^2}]\] (Again using identities mentioned in Hint number 3 and 4)
\[ = 2{\tan ^2}\theta - 2{\cot ^2}\theta - [1 + {\tan ^4}\theta + 2{\tan ^2}\theta ] + [1 + {\cot ^4}\theta + 2{\cot ^2}\theta ]\]
Using identity ${(a + b)^2} = {a^2} + {b^2} + 2ab$
\[ = 2{\tan ^2}\theta - 2{\cot ^2}\theta - 1 - {\tan ^4}\theta - 2{\tan ^2}\theta + 1 + {\cot ^4}\theta + 2{\cot ^2}\theta \]
Cancelling alike and opposite terms we will get the required answer
\[ = {\cot ^4}\theta - {\tan ^4}\theta \]
Hence proved.
Note:
1. This question can be done by various methods one method is by converting the whole equation into $\sin \theta $ and $\cos \theta $. But it will be very long and time consuming. The approach by which we have solved is way more convenient rather than converting the whole into $\sin \theta $ and $\cos \theta $. So some identities need to be memorised so that it can be done easily and fast.
2. Be very cautious when opening the degree of a trigonometric function. For example: - ${\sec ^4}\theta = {\left( {{{\sec }^2}\theta } \right)^2}$
Importance of trigonometric functions: - In geometry trigonometric functions are used to find unknown angles or sides of right angled triangles. The three common used trigonometric functions are $\sin \theta ,\cos \theta ,\tan \theta $.If we have good understanding of these three functions then other trigonometric functions such as $\cos ec\theta ,\sec \theta ,\cot \theta $ which are reciprocal of the above three functions can b easily understood.
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