How do you simplify \[{\sec ^4}x - {\tan ^4}x\]
Answer
629.4k+ 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.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

