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Question: How do you evaluate \({{\sec }^{2}}\left( 18 \right)-{{\tan }^{2}}\left( 18 \right)\) ?...

How do you evaluate sec⁡2(18)−tan⁡2(18){{\sec }^{2}}\left( 18 \right)-{{\tan }^{2}}\left( 18 \right) ?

Explanation

Solution

Whenever complex equations are given to solve one must always Firstly start from the complex side and then convert all the terms into cos⁡θ\cos \theta or sin⁡θ\sin \theta . So here we convert sec⁡θ\sec \theta into 1cos⁡θ\dfrac{1}{\cos \theta } and tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } and then combine them into single fractions and then evaluate to find the value of the expressions.

Complete step by step solution:
The given trigonometric expression, sec⁡2(18)−tan⁡2(18){{\sec }^{2}}\left( 18 \right)-{{\tan }^{2}}\left( 18 \right)
Now convert the above expression entirely in terms of cos⁡θ\cos \theta or sin⁡θ\sin \theta .
Here according to our question, θ=18\theta =18
We convert it into basic trigonometric functions so that we can evaluate it easily.
Let us now convert the first term of the expression.
Since sec⁡θ\sec \theta is 1cos⁡θ\dfrac{1}{\cos \theta } , solving the given trigonometric quantity using cos⁡θ\cos \thetawould be easier.
After converting we get,
⇒1cos⁡2(18)−tan⁡2(18)\Rightarrow \dfrac{1}{{{\cos }^{2}}\left( 18 \right)}-{{\tan }^{2}}\left( 18 \right)
Now let us convert the second term of the expression.
Since tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } substitute it in our expression.
Here according to our question, θ=18\theta =18
After converting we get,
⇒1cos⁡2(18)−sin⁡2(18)cos⁡2(18)\Rightarrow \dfrac{1}{{{\cos }^{2}}\left( 18 \right)}-\dfrac{{{\sin }^{2}}\left( 18 \right)}{{{\cos }^{2}}\left( 18 \right)}
Now let us evaluate.
Since the denominators are the same, we shall now subtract the numerators.
⇒1−sin⁡2(18)cos⁡2(18)\Rightarrow \dfrac{1-{{\sin }^{2}}\left( 18 \right)}{{{\cos }^{2}}\left( 18 \right)}
Now know the basic trigonometric postulate sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
Since our numerator of the expression in the same format we,
Substituting 1−sin⁡2θ=cos⁡2θ1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta
Here according to our question, θ=18\theta =18
On evaluating We get,
⇒cos⁡2(18)cos⁡2(18)\Rightarrow \dfrac{{{\cos }^{2}}\left( 18 \right)}{{{\cos }^{2}}\left( 18 \right)}
Since the numerator and the denominator are the same, we can cancel the common terms.
On evaluating we get,
⇒cos⁡2(18)cos⁡2(18)=1\Rightarrow \dfrac{{{\cos }^{2}}\left( 18 \right)}{{{\cos }^{2}}\left( 18 \right)}=1
Hence sec⁡2(18)−tan⁡2(18){{\sec }^{2}}\left( 18 \right)-{{\tan }^{2}}\left( 18 \right) is equal to 11

Note: It is a must to memorize the values of basic trigonometric functions since all the functions can be written in terms of those basic trigonometric functions and can be easily evaluated. Always check when the trigonometric functions are given in degrees or radians. There is a lot of difference between both 1∘×π180=0.017Rad1{}^\circ \times \dfrac{\pi }{180}=0.017Rad .