
If in a triangle ABC, angle C is \[45^\circ \], then \[\left( {1 + cotA} \right)\left( {1 + cotB} \right)\]=
A. \[ - 1\]
B. \[2\]
C. \[3\]
D. \[\sqrt {\frac{1}{2}} \]
Answer
216k+ views
Hint
To solve this problem, we need to use the law of sines. The law of sines states that in a right triangle, the sum of the two angles is equal to\[120^\circ \]. So, \[\left( {1 + cotA} \right)\left( {1 + cotB} \right) = 120^\circ \].
According to the law of sine, or sine law, the ratio of a triangle's side length to the sine of the opposing angle remains constant for all three sides. The sine rule is another name for it. The relationship between the sides and angles of non-right (oblique) triangles is known as the Law of Sines. It simply asserts that for all sides and angles of a given triangle, the ratio of the length of a side to the sine of the angle opposite that side is the same.
Complete step-by-step solution:
The given equation is\[\angle C = 45^\circ \]and
\[A + B = 180^\circ - 45^\circ = 135^\circ \]
\[\cot (A + B) = \frac{{\cot A \cot B - 1}}{{\cot A + \cot B}}\]
As the value of A+B is already known, the equation becomes,
\[\cot (135^\circ ) = - 1 = \frac{{\cot A \cot B - 1}}{{\cot A + \cot B}}\]
\[\cot A + \cot B = 1 - \cot A\cot B\]
\[1 + \cot A + \cot B + \cot A\cot B\]
Then, the answer is equal to \[2\]
\[(1 + \cot A)(1 + \cot B) = 2\]
So, the option B is correct.
Note
Use an ordering of operations to solve the triangle issue. The parenthesis is used to emphasize that the contents should be considered first before anything else. Since angle C is \[45\] degrees in this instance, it is analyzed first, leading to the results: \[\left( {1 + cotA} \right)\left( {1 + cotB} \right) = 2{\rm{ }}or{\rm{ }}1 + \left( {cot\left( {45} \right)} \right) = \left( {1 + cotB} \right)\].
A phrase, word, or sentence added to writing as supplemental information or an afterthought is known as a parenthesis. It is emphasized with bracketed text, commas, or dashes.
Multiplying numbers is the second application of parentheses in mathematics. When parenthesis is used in an equation and there is no arithmetic operation, multiplication must be used.
To solve this problem, we need to use the law of sines. The law of sines states that in a right triangle, the sum of the two angles is equal to\[120^\circ \]. So, \[\left( {1 + cotA} \right)\left( {1 + cotB} \right) = 120^\circ \].
According to the law of sine, or sine law, the ratio of a triangle's side length to the sine of the opposing angle remains constant for all three sides. The sine rule is another name for it. The relationship between the sides and angles of non-right (oblique) triangles is known as the Law of Sines. It simply asserts that for all sides and angles of a given triangle, the ratio of the length of a side to the sine of the angle opposite that side is the same.
Complete step-by-step solution:
The given equation is\[\angle C = 45^\circ \]and
\[A + B = 180^\circ - 45^\circ = 135^\circ \]
\[\cot (A + B) = \frac{{\cot A \cot B - 1}}{{\cot A + \cot B}}\]
As the value of A+B is already known, the equation becomes,
\[\cot (135^\circ ) = - 1 = \frac{{\cot A \cot B - 1}}{{\cot A + \cot B}}\]
\[\cot A + \cot B = 1 - \cot A\cot B\]
\[1 + \cot A + \cot B + \cot A\cot B\]
Then, the answer is equal to \[2\]
\[(1 + \cot A)(1 + \cot B) = 2\]
So, the option B is correct.
Note
Use an ordering of operations to solve the triangle issue. The parenthesis is used to emphasize that the contents should be considered first before anything else. Since angle C is \[45\] degrees in this instance, it is analyzed first, leading to the results: \[\left( {1 + cotA} \right)\left( {1 + cotB} \right) = 2{\rm{ }}or{\rm{ }}1 + \left( {cot\left( {45} \right)} \right) = \left( {1 + cotB} \right)\].
A phrase, word, or sentence added to writing as supplemental information or an afterthought is known as a parenthesis. It is emphasized with bracketed text, commas, or dashes.
Multiplying numbers is the second application of parentheses in mathematics. When parenthesis is used in an equation and there is no arithmetic operation, multiplication must be used.
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