
If \[\sin (A + B) = 1\] and\[\cos (A - B) = \frac{{\sqrt 3 }}{2}\], then the smallest positive values of\[{\rm{A}}\]and\[{\rm{B}}\]are
A. \[{60^\circ },{30^\circ }\]
В. \[{75^\circ },{15^\circ }\]
C. \[{45^\circ },{60^\circ }\]
D. \[{45^\circ },{45^\circ }\]
Answer
163.2k+ views
Hint: An equation based on trigonometry is said to be trigonometry identity which is always true. The sum and differences can be determined using \[sin(\alpha + \beta ) = sin\alpha cos\beta + cos\alpha sin\beta \]and\[sin(\alpha - \beta ) = sin\alpha cos\beta - cos\alpha sin\beta \].In this case\[\sin (A + B) = 1\] and\[\cos (A - B) = \frac{{\sqrt 3 }}{2}\], the smallest positive value can be obtained by solving them using trigonometry identities and sum and differences of two angles concept.
Complete step by step solution:We have given the equation,
\[\sin (A + B) = 1\]
This can be written as
\[\sin (A + B) = \sin {90^ \circ } \Rightarrow A + B = {90^ \circ }\]-- (i)
And\[\cos (A - B) = \frac{{\sqrt 3 }}{2}\]
\[ \Rightarrow \cos (A - B) = \cos {30^\circ } \Rightarrow A - B = {30^\circ }\]-- (ii)
Adding the resultant equation (i) and (ii)
The equations are,
\[A + B = {90^\circ }\]
\[A - B = {30^\circ }\]
Add the equations and cancel the similar terms:
\[A + B + A - B = {90^\circ } + {60^\circ }\]
Group the like terms and add:
\[2A = {120^\circ }\]
Divide either side of the equation by\[2\]and solve for A:
\[ \Rightarrow A = {60^\circ }\]
Put\[A = {60^\circ }\]in equation (1)
\[{60^\circ } + B = {90^\circ }\]
That implies:
Subtracting equation (ii) from (i)
\[ \Rightarrow B = {90^\circ } - {60^\circ } = {30^\circ }\]
Hence, we obtain the value as
\[\therefore B = {30^\circ }\]
So, the values are
\[ \Rightarrow A = {60^\circ },B = {30^\circ }\]
This can be also written as,
\[A = \frac{\pi }{3},B = \frac{\pi }{6}\]
Hence, the smallest positive values of\[{\rm{A}}\]and\[{\rm{B}}\]are\[{60^\circ },{30^\circ }\]
Option ‘A’ is correct
Note: Although the question initially appeared to be somewhat large, it turns out to be fairly simple to answer using a trigonometric formula. So, in order to solve this kind of problem, we must learn the key formula, and we can only accomplish this by practice. Therefore, it is important to understand how to use trigonometric identities and to build identities in the correct order. It simplifies the problem and aids in your quest for the right response. This method makes it very simple for us to solve this kind of issue.
Complete step by step solution:We have given the equation,
\[\sin (A + B) = 1\]
This can be written as
\[\sin (A + B) = \sin {90^ \circ } \Rightarrow A + B = {90^ \circ }\]-- (i)
And\[\cos (A - B) = \frac{{\sqrt 3 }}{2}\]
\[ \Rightarrow \cos (A - B) = \cos {30^\circ } \Rightarrow A - B = {30^\circ }\]-- (ii)
Adding the resultant equation (i) and (ii)
The equations are,
\[A + B = {90^\circ }\]
\[A - B = {30^\circ }\]
Add the equations and cancel the similar terms:
\[A + B + A - B = {90^\circ } + {60^\circ }\]
Group the like terms and add:
\[2A = {120^\circ }\]
Divide either side of the equation by\[2\]and solve for A:
\[ \Rightarrow A = {60^\circ }\]
Put\[A = {60^\circ }\]in equation (1)
\[{60^\circ } + B = {90^\circ }\]
That implies:
Subtracting equation (ii) from (i)
\[ \Rightarrow B = {90^\circ } - {60^\circ } = {30^\circ }\]
Hence, we obtain the value as
\[\therefore B = {30^\circ }\]
So, the values are
\[ \Rightarrow A = {60^\circ },B = {30^\circ }\]
This can be also written as,
\[A = \frac{\pi }{3},B = \frac{\pi }{6}\]
Hence, the smallest positive values of\[{\rm{A}}\]and\[{\rm{B}}\]are\[{60^\circ },{30^\circ }\]
Option ‘A’ is correct
Note: Although the question initially appeared to be somewhat large, it turns out to be fairly simple to answer using a trigonometric formula. So, in order to solve this kind of problem, we must learn the key formula, and we can only accomplish this by practice. Therefore, it is important to understand how to use trigonometric identities and to build identities in the correct order. It simplifies the problem and aids in your quest for the right response. This method makes it very simple for us to solve this kind of issue.
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