Question

If\\[\cos \theta = - \frac{1}{{\sqrt 2 }}\] \and\\[\tan \theta = 1\], then the general value of\[\theta \]\\is
A. \[2n\pi + \frac{\pi }{4}\]
В. \[(2n + 1)\pi + \frac{\pi }{4}\]
C. \[n\pi + \frac{\pi }{4}\]
D. \[n\pi \pm \frac{\pi }{4}\]

Answer 1

Hint: Sometimes anything will become a trig identity by factoring with a widely used phrase. Multiply the conjugate by the denominator. Find the least common factor. A fraction can be divided into two smaller fractions. Everything should be rewritten using sine and cosine. Continue by utilizing trigonometric identities to solve the equation. In this case, \[\cos \theta = - \frac{1}{{\sqrt 2 }}\]and\[\tan \theta = 1\], first step is that you can factor. In order to recall this, we must be aware of the fundamental coordinate system used by all trigonometric functions as well as their general solution formula.




Complete step by step solution:We have given the equation according to the question:
\[\cos \theta = - \frac{1}{{\sqrt 2 }}\]and\\[\tan \theta = 1\]
Rewrite the equation given in terms of\[cos\]:
\[\cos \theta = - 1/\sqrt 2 = - \cos (\pi /4)\]
Solve to obtain the required values in order to find the value of theta:
\[ = \cos (\pi - \pi /4){\rm{ or }}\cos (\pi + \pi /4)\]
Determine the value for\[\theta \]:
\[\therefore \theta = 3\pi /4,5\pi /4\]
Now solve the equation in terms of tan:
\[\tan \theta = 1 = \tan (\pi /4),\tan (\pi + \pi /4)\]
Determine the value for\[\theta \]:
\[\therefore \theta = \pi /4,5\pi /4\]
Hence the value of\[\theta \]between\[0\]and \[2\pi \]which satisfies both the equations is\[5\pi /4\]
\[\therefore \]General value of the given equation is\[2n\pi + 5\pi /4 = (2{\rm{n}} + 1)\pi + \pi /4\].
Hence, the general value of\[\theta \]is\[(2n + 1)\pi + \frac{\pi }{4}\]



Option ‘B’ is correct

Note: Student makes errors while applying identities. Not able to comprehend how to arrive at the other side Students at times apply identities incorrectly \[sin\theta = 1 - cos\theta \]or\[tan\theta = 1 + {\rm{ }}sec\theta \]. Start the simplification of the equation, on both the sides simultaneously.
Due to the fact that this value is repeated after every n interval, we can only choose one common value of theta. And utilizing graphical techniques, we can discover this.