Question

If\[(1 + \tan \theta )(1 + \tan \varphi ) = 2\], then\[\theta + \varphi = \]
A. \[{30^\circ }\]
B. \[{45^\circ }\]
C. \[{60^\circ }\]
D. \[{75^\circ }\]

Answer 1

Hint: In this case\[(1 + \tan \theta )(1 + \tan \varphi ) = 2\], first simply expand the given equation and cancel the common terms and solve the rest using trigonometry identities. As the value of theta is known as\[\theta = {45^\circ }\], it can be solved accordingly. Since it is difficult to predict what to add to the trigonometric equation, it is more likely to start by roughly solving the offered possibilities in order to comprehend the options that give the statement. We then calculate the additional expression using a backwards technique.



Formula Used:

Complete step by step solution:We have given the equation, according to the question is
\[(1 + \tan \theta )(1 + \tan \phi ) = 2\]
Rewrite as:
\[ \Rightarrow 1 + \tan \theta + \tan \phi + \tan \theta \tan \phi = 2\]
Subtract\[2\]from both sides:
\[\tan \left( \phi \right)\tan \left( \theta \right) + \tan \left( \theta \right) + \tan \left( \phi \right) - 1 = 0\]
Can be written as:
\[ \Rightarrow \tan \theta + \tan \phi = 1 - \tan \theta \tan \phi \]
Express with\[\sin \]\\[\cos \]:
\[\dfrac{{ - \cos (\varphi )\cos (\theta ) + \cos (\varphi )\sin (\theta ) + \cos (\theta )\sin (\varphi ) + \sin (\varphi )\sin (\theta )}}{{\cos (\varphi )\cos (\theta )}} = 0\]
By\[\dfrac{{f(x)}}{{g(x)}} = 0 \Rightarrow f(x) = 0\]:
\[ - \cos (\varphi )\cos (\theta ) + \cos (\varphi )\sin (\theta ) + \cos (\theta )\sin (\varphi ) + \sin (\varphi )\sin (\theta ) = 0\]
Rewrite the above using trig identities:
\[\sqrt 2 \sin \left( {\varphi + \theta - \dfrac{\pi }{4}} \right) = 0\]
Divide either side by\[\sqrt 2 \]:
\[\dfrac{{\sqrt 2 \sin \left( {\varphi + \theta - \dfrac{\pi }{4}} \right)}}{{\sqrt 2 }} = \dfrac{0}{{\sqrt 2 }}\]
Simplify the above equation so as to obtain the less complicated form:
\[\sin \left( {\varphi + \theta - \dfrac{\pi }{4}} \right) = 0\]
General solutions for\[\sin \left( {\varphi + \theta - \dfrac{\pi }{4}} \right) = 0\]:
\[\varphi + \theta - \frac{\pi }{4} = 0 + 2\pi n,\varphi + \theta - \frac{\pi }{4} = \pi + 2\pi n\]
Solve\[\varphi + \theta - \frac{\pi }{4} = 0 + 2\pi n:\]
\[\theta = 2\pi n - \varphi + \frac{\pi }{4}\]
Solve\[\varphi + \theta - \frac{\pi }{4} = \pi + 2\pi n:\]
\[\theta = \pi + 2\pi n - \varphi + \frac{\pi }{4}\]
Can be written as:
\[\theta = 2\pi n - \varphi + \dfrac{\pi }{4},\theta = \pi + 2\pi n - \varphi + \frac{\pi }{4}\]
Rewrite in terms of\[\tan \]:
\[ \Rightarrow \dfrac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }} = 1\]
Solve the fraction to make it to simpler form:
\[ \Rightarrow \tan (\theta + \phi ) = 1\]
Simplify using trigonometry formula:
\[ \Rightarrow \theta + \phi = {45^\circ }\]
Hence, the value of \[\theta + \phi = {45^\circ }\].



Option ‘B’ is correct

Note: Sometimes students make error in applying identities and are not able to comprehend how to arrive at the other. It is to known that the ratio of the opponent to the adjacent side is known as the tangent. Students occasionally mistakenly apply identities. Making the mistake of simplifying both sides simultaneously, when applying trigonometric identities is a typical error. Do not always try to solve the question only through trigonometric equations and functions.