Answer
Verified
482.4k+ views
Hint- Use the algebraic identity \[\left( {{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)} \right)\] and \[\left( {{{\sec }^2}\theta - {{\tan }^{^2}}\theta } \right) = 1\].
According to question Let first rewrite the equation given as below-
\[{\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta = 1\]
Taking the LHS and rearrange the equation as below
\[{\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta \]
\[ \Rightarrow {\left( {{{\sec }^2}\theta } \right)^3} - {\left( {{{\tan }^2}\theta } \right)^3} - 3{\sec ^2}\theta {\tan ^2}\theta \] ………. (1)
Now if we clearly look the above equation it is of the form \[{a^3} - {b^3}\] and by algebraic identity we know that \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\]
So rewriting the equation (1) we get
\[ \Rightarrow \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\left( {{{\sec }^{^4}}\theta + {{\tan }^4}\theta + {{\sec }^2}\theta {{\tan }^2}\theta } \right) - 3{\sec ^2}\theta {\tan ^{^2}}\theta \]
Now by trigonometric identity we know that \[\left\{ {{{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right\}\]
\[ \Rightarrow 1\left( {{{\sec }^{^4}}\theta + {{\tan }^4}\theta + {{\sec }^2}\theta {{\tan }^2}\theta } \right) - 3{\sec ^2}\theta {\tan ^{^2}}\theta \]
\[ \Rightarrow {\sec ^{^4}}\theta + {\tan ^4}\theta + {\sec ^2}\theta {\tan ^2}\theta - 3{\sec ^2}\theta {\tan ^{^2}}\theta \]
\[ \Rightarrow {\sec ^{^4}}\theta + {\tan ^4}\theta - 2{\sec ^2}\theta {\tan ^{^2}}\theta \]
Or, above can be written as
\[ \Rightarrow {\left( {{{\sec }^2}\theta - {{\tan }^{^2}}\theta } \right)^2}{\left( {{{\tan }^{^2}}\theta - {{\sec }^2}\theta } \right)^2}\]
Again, by trigonometric identity we know that \[\left\{ {{{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right\}\]
\[\
\Rightarrow {\left( 1 \right)^2}{\text{ or }}{\left( { - 1} \right)^2} \\
\Rightarrow 1 \\
\]
Therefore LHS \[{\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta = 1\] or we can say\[{\sec ^6}\theta = 1 + {\tan ^{^6}}\theta + 3{\sec ^2}\theta {\tan ^2}\theta \]
Hence Proved.
Note- Whenever this type of question appears always bring the RHS terms to LHS and then solve LHS. Afterwards rearrange the equation such that it makes an algebraic identity [ as in our question we converted the equation \[{\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta \] to the form \[{a^3} - {b^3}\]].Remember the trigonometric identity \[\left( {{{\sec }^2}\theta - {{\tan }^{^2}}\theta } \right) = 1\].
According to question Let first rewrite the equation given as below-
\[{\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta = 1\]
Taking the LHS and rearrange the equation as below
\[{\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta \]
\[ \Rightarrow {\left( {{{\sec }^2}\theta } \right)^3} - {\left( {{{\tan }^2}\theta } \right)^3} - 3{\sec ^2}\theta {\tan ^2}\theta \] ………. (1)
Now if we clearly look the above equation it is of the form \[{a^3} - {b^3}\] and by algebraic identity we know that \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\]
So rewriting the equation (1) we get
\[ \Rightarrow \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\left( {{{\sec }^{^4}}\theta + {{\tan }^4}\theta + {{\sec }^2}\theta {{\tan }^2}\theta } \right) - 3{\sec ^2}\theta {\tan ^{^2}}\theta \]
Now by trigonometric identity we know that \[\left\{ {{{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right\}\]
\[ \Rightarrow 1\left( {{{\sec }^{^4}}\theta + {{\tan }^4}\theta + {{\sec }^2}\theta {{\tan }^2}\theta } \right) - 3{\sec ^2}\theta {\tan ^{^2}}\theta \]
\[ \Rightarrow {\sec ^{^4}}\theta + {\tan ^4}\theta + {\sec ^2}\theta {\tan ^2}\theta - 3{\sec ^2}\theta {\tan ^{^2}}\theta \]
\[ \Rightarrow {\sec ^{^4}}\theta + {\tan ^4}\theta - 2{\sec ^2}\theta {\tan ^{^2}}\theta \]
Or, above can be written as
\[ \Rightarrow {\left( {{{\sec }^2}\theta - {{\tan }^{^2}}\theta } \right)^2}{\left( {{{\tan }^{^2}}\theta - {{\sec }^2}\theta } \right)^2}\]
Again, by trigonometric identity we know that \[\left\{ {{{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right\}\]
\[\
\Rightarrow {\left( 1 \right)^2}{\text{ or }}{\left( { - 1} \right)^2} \\
\Rightarrow 1 \\
\]
Therefore LHS \[{\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta = 1\] or we can say\[{\sec ^6}\theta = 1 + {\tan ^{^6}}\theta + 3{\sec ^2}\theta {\tan ^2}\theta \]
Hence Proved.
Note- Whenever this type of question appears always bring the RHS terms to LHS and then solve LHS. Afterwards rearrange the equation such that it makes an algebraic identity [ as in our question we converted the equation \[{\sec ^6}\theta - {\tan ^{^6}}\theta - 3{\sec ^2}\theta {\tan ^2}\theta \] to the form \[{a^3} - {b^3}\]].Remember the trigonometric identity \[\left( {{{\sec }^2}\theta - {{\tan }^{^2}}\theta } \right) = 1\].
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE