
How do you simplify \[{\sec ^4}x - {\tan ^4}x\]
Answer
561k+ views
Hint:The question is related to the trigonometry. By using the property of trigonometry identities and some formulas based on the algebraic we are going to simplify the given question. The trigonometry has its ratios and some trigonometry ratios are interlinked to the other trigonometry ratios.
Complete step by step explanation: Here in this question the tan and secant are the trigonometry ratios.
The identity in the question is given as \[{\sec ^4}x - {\tan ^4}x\] , this can be written as
\[ \Rightarrow {\left( {{{\sec }^2}x} \right)^2} - {\left( {{{\tan }^2}x} \right)^2}\] ------ (1)
The above inequality is in the form of the standard algebraic formula that is \[{a^2} - {b^2} = (a - b)(a
+ b)\]
By using this standard algebraic formula, the above inequality can be written as
\[ \Rightarrow \left( {{{\sec }^2}x - {{\tan }^2}x} \right)\left( {{{\sec }^2}x + {{\tan }^2}x} \right)\] ------- (2)
Now we will consider the \[\left( {{{\sec }^2}x - {{\tan }^2}x} \right)\] , we apply the trigonometry ratios. The \[\sec x\] is also written as \[\dfrac{1}{{\cos x}}\] and the \[\tan x\] is also written as
\[\dfrac{{\sin x}}{{\cos x}}\]
By using the above trigonometry ratios, we have
\[\left( {{{\sec }^2}x - {{\tan }^2}x} \right) = \dfrac{1}{{{{\cos }^2}x}} - \dfrac{{{{\sin }^2}x}}{{{{\cos
}^2}x}}\] -------- (3)
Since the denominator of the both terms are same the above equation is written as
\[ \Rightarrow \left( {{{\sec }^2}x - {{\tan }^2}x} \right) = \dfrac{{1 - {{\sin }^2}x}}{{{{\cos }^2}x}}\] --------- (4)
We know the trigonometry identity \[{\sin ^2}x + {\cos ^2}x = 1\]
By shifting \[{\sin ^2}x\] to the RHS the identity is written as \[1 - {\sin ^2}x = {\cos ^2}x\] ---- (5)
Substitute equation (5) in equation (4) we have
\[ \Rightarrow \left( {{{\sec }^2}x - {{\tan }^2}x} \right) = \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}}\]
Since the both numerator and denominator having the same value we can cancel, so we have
\[ \Rightarrow \left( {{{\sec }^2}x - {{\tan }^2}x} \right) = 1\] ----- (6)
Substitute the equation (6) in equation (2) we have
\[ \Rightarrow 1 \cdot \left( {{{\sec }^2}x + {{\tan }^2}x} \right)\] ------ (7)
We know the trigonometry identity \[{\sec ^2}x = 1 + {\tan ^2}x\] ------ (8)
Substitute the equation (8) in the equation (7) we have
\[ \Rightarrow 1 + {\tan ^2}x + {\tan ^2}x\]
We can add the similar terms, therefore we have
\[ \Rightarrow 1 + 2{\tan ^2}x\]
Hence, we obtained the solution as \[1 + 2{\tan ^2}x\] by simplifying the \[{\sec ^4}x - {\tan ^4}x\]
Note:In the trigonometry the trigonometry ratios and trigonometry identities are used to simplify the given question. The trigonometry ratios are interlinked to other trigonometry ratios. The standard algebraic formulas are used wherever it is used or necessary.
Complete step by step explanation: Here in this question the tan and secant are the trigonometry ratios.
The identity in the question is given as \[{\sec ^4}x - {\tan ^4}x\] , this can be written as
\[ \Rightarrow {\left( {{{\sec }^2}x} \right)^2} - {\left( {{{\tan }^2}x} \right)^2}\] ------ (1)
The above inequality is in the form of the standard algebraic formula that is \[{a^2} - {b^2} = (a - b)(a
+ b)\]
By using this standard algebraic formula, the above inequality can be written as
\[ \Rightarrow \left( {{{\sec }^2}x - {{\tan }^2}x} \right)\left( {{{\sec }^2}x + {{\tan }^2}x} \right)\] ------- (2)
Now we will consider the \[\left( {{{\sec }^2}x - {{\tan }^2}x} \right)\] , we apply the trigonometry ratios. The \[\sec x\] is also written as \[\dfrac{1}{{\cos x}}\] and the \[\tan x\] is also written as
\[\dfrac{{\sin x}}{{\cos x}}\]
By using the above trigonometry ratios, we have
\[\left( {{{\sec }^2}x - {{\tan }^2}x} \right) = \dfrac{1}{{{{\cos }^2}x}} - \dfrac{{{{\sin }^2}x}}{{{{\cos
}^2}x}}\] -------- (3)
Since the denominator of the both terms are same the above equation is written as
\[ \Rightarrow \left( {{{\sec }^2}x - {{\tan }^2}x} \right) = \dfrac{{1 - {{\sin }^2}x}}{{{{\cos }^2}x}}\] --------- (4)
We know the trigonometry identity \[{\sin ^2}x + {\cos ^2}x = 1\]
By shifting \[{\sin ^2}x\] to the RHS the identity is written as \[1 - {\sin ^2}x = {\cos ^2}x\] ---- (5)
Substitute equation (5) in equation (4) we have
\[ \Rightarrow \left( {{{\sec }^2}x - {{\tan }^2}x} \right) = \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}}\]
Since the both numerator and denominator having the same value we can cancel, so we have
\[ \Rightarrow \left( {{{\sec }^2}x - {{\tan }^2}x} \right) = 1\] ----- (6)
Substitute the equation (6) in equation (2) we have
\[ \Rightarrow 1 \cdot \left( {{{\sec }^2}x + {{\tan }^2}x} \right)\] ------ (7)
We know the trigonometry identity \[{\sec ^2}x = 1 + {\tan ^2}x\] ------ (8)
Substitute the equation (8) in the equation (7) we have
\[ \Rightarrow 1 + {\tan ^2}x + {\tan ^2}x\]
We can add the similar terms, therefore we have
\[ \Rightarrow 1 + 2{\tan ^2}x\]
Hence, we obtained the solution as \[1 + 2{\tan ^2}x\] by simplifying the \[{\sec ^4}x - {\tan ^4}x\]
Note:In the trigonometry the trigonometry ratios and trigonometry identities are used to simplify the given question. The trigonometry ratios are interlinked to other trigonometry ratios. The standard algebraic formulas are used wherever it is used or necessary.
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