If sec θ – tan θ= x, show that : sec θ = 1 2 (x + 1 x ) and tan θ = 1 2 ( 1 x − x)
Answers
Answer:
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Step-by-step explanation:
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Step-by-step explanation:
Answer: Showed
Step-by-step explanation: Given relation is as follows
\sec \theta-\tan \theta=x.secθ−tanθ=x.
We are given to prove the following two results-
\sec \theta=\dfrac{1}{2}(x+\dfrac{1}{x})secθ=
2
1
(x+
x
1
) and \tan \theta=\dfrac{1}{2}(\dfrac{1}{x}-x).tanθ=
2
1
(
x
1
−x).
To prove the above two equalities, we will need the following relation from 'Trigonometry'-
\begin{gathered}\sec^{2}\theta=1+\tan^{2}\theta\\\\\Rightarrow \sec^2\theta-\tan^2\theta=1.\end{gathered}
sec
2
θ=1+tan
2
θ
⇒sec
2
θ−tan
2
θ=1.
Now,
\begin{gathered}\sec \theta=\dfrac{1}{2}(\sec \theta-\tan \theta+\sec \theta+\tan \theta)\\\\\Rightarrow \sec \theta=\dfrac{1}{2}(\sec \theta-\tan \theta+\dfrac{\sec \theta+\tan \theta}{\sec^2\theta-\tan^2\theta})\\\\\Rightarrow \sec \theta=\dfrac{1}{2}(\sec \theta-\tan \theta+\dfrac{1}{\sec \theta-\tan \theta})\\\\\Rightarrow \sec \theta=\dfrac{1}{2}(x+\dfrac{1}{x}),\end{gathered}
secθ=
2
1
(secθ−tanθ+secθ+tanθ)
⇒secθ=
2
1
(secθ−tanθ+
sec
2
θ−tan
2
θ
secθ+tanθ
)
⇒secθ=
2
1
(secθ−tanθ+
secθ−tanθ
1
)
⇒secθ=
2
1
(x+
x
1
),
and
\begin{gathered}\tan \theta=\dfrac{1}{2}(\sec \theta+\tan \theta-\sec \theta+\tan \theta)\\\\\Rightarrow \tan \theta=\dfrac{1}{2}(\dfrac{\sec \theta+\tan \theta}{\sec^2\theta-\tan^2\theta}-\sec \theta+\tan \theta)\\\\\Rightarrow \tan \theta=\dfrac{1}{2}(\dfrac{1}{\sec \theta-\tan \theta}-(\sec \theta-\tan \theta))\\\\\Rightarrow \tan \theta=\dfrac{1}{2}(\dfrac{1}{x}-x).\end{gathered}
tanθ=
2
1
(secθ+tanθ−secθ+tanθ)
⇒tanθ=
2
1
(
sec
2
θ−tan
2
θ
secθ+tanθ
−secθ+tanθ)
⇒tanθ=
2
1
(
secθ−tanθ
1
−(secθ−tanθ))
⇒tanθ=
2
1
(
x
1
−x).