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Question: Prove that \({{\sec }^{4}}\theta -{{\sec }^{2}}\theta ={{\tan }^{4}}\theta +{{\tan }^{2}}\theta \)....

Prove that sec⁡4θ−sec⁡2θ=tan⁡4θ+tan⁡2θ{{\sec }^{4}}\theta -{{\sec }^{2}}\theta ={{\tan }^{4}}\theta +{{\tan }^{2}}\theta .

Explanation

Solution

Hint: Take sec⁡2θ{{\sec }^{2}}\theta common from the terms in L.H.S. and use sec⁡2θ=1+tan⁡2θ{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta and simplify the expression. Hence prove that L.H.S. = R.H.S.

Complete step-by-step answer:
Trigonometric ratios:
There are six trigonometric ratios defined on an angle of a right-angled triangle, viz sine, cosine, tangent, cotangent, secant and cosecant.
The sine of an angle is defined as the ratio of the opposite side to the hypotenuse.
The cosine of an angle is defined as the ratio of the adjacent side to the hypotenuse.
The tangent of an angle is defined as the ratio of the opposite side to the adjacent side.
The cotangent of an angle is defined as the ratio of the adjacent side to the opposite side.
The secant of an angle is defined as the ratio of the hypotenuse to the adjacent side.
The cosecant of an angle is defined as the ratio of the hypotenuse to the opposite side.
Pythagorean identities:
The identities sin⁡2θ+cos⁡2θ=1,sec⁡2θ=1+tan⁡2θ{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta and csc⁡2θ=1+cot⁡2θ{{\csc }^{2}}\theta =1+{{\cot }^{2}}\theta are known as Pythagorean identities as they are a direct consequence of Pythagoras’ theorem in a right-angled triangle.
We have L.H.S. =sec⁡4θ−sec⁡2θ={{\sec }^{4}}\theta -{{\sec }^{2}}\theta
We know that sec⁡2θ=1+tan⁡2θ{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta .
Hence, we have
L.H.S. =(1+tan⁡2θ)2−(1+tan⁡2θ)={{\left( 1+{{\tan }^{2}}\theta \right)}^{2}}-\left( 1+{{\tan }^{2}}\theta \right)
We know that (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}
Using the above algebraic identity, we get
L.H.S. =1+2tan⁡2θ+tan⁡4θ−1−tan⁡2θ=1+2{{\tan }^{2}}\theta +{{\tan }^{4}}\theta -1-{{\tan }^{2}}\theta
Hence, we have L.H.S. =tan⁡4θ+tan⁡2θ={{\tan }^{4}}\theta +{{\tan }^{2}}\theta
Hence, we have L.H.S. = R.H.S.

Note: Alternative solution.
We have L.H.S =sec⁡4θ−sec⁡2θ={{\sec }^{4}}\theta -{{\sec }^{2}}\theta
Taking sec⁡2θ{{\sec }^{2}}\theta common, we get
L.H.S =sec⁡2θ(sec⁡2θ−1)={{\sec }^{2}}\theta \left( {{\sec }^{2}}\theta -1 \right)
Using sec⁡2θ=1+tan⁡2θ{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta , we get
L.H.S =(1+tan⁡2θ)(tan⁡2θ)=\left( 1+{{\tan }^{2}}\theta \right)\left( {{\tan }^{2}}\theta \right)
Using the distributive property of multiplication over addition, i.e. a(b+c) = ab+ac, we get
L.H.S =tan⁡2θ+tan⁡4θ={{\tan }^{2}}\theta +{{\tan }^{4}}\theta
Hence, we have L.H.S = R.H.S