
If \[co{s^{ - 1}}\sqrt p + co{s^{ - 1}}\sqrt {\left( {1{\text{ }} - {\text{ }}p} \right)} + co{s^{ - 1}}\sqrt {\left( {1{\text{ }} - {\text{ }}q} \right)} = \dfrac{{3\pi }}{4}\], then the value of \[q\] is
A. \[1\]
B. \[\dfrac{1}{{\sqrt 2 }}\]
C. \[\dfrac{1}{3}\]
D. \[\dfrac{1}{2}\]
Answer
232.8k+ views
Hint: Use the formulas of inverse trigonometric functions to solve the given question. Here we convert the second cos inverse ratio to a sin inverse ratio by using a formula for simplification. By substituting the formulas and simplifying we can obtain the value of q.
Formula Used:
\[{\sin ^{ - 1}}x = {\cos ^{ - 1}}\sqrt {1 - {x^2}} \]
\[{\cos^{ - 1}}x + si{n^{ - 1}}x = \dfrac{\pi }{2}\]
Complete step by step Solution:
\[co{s^{ - 1}}\sqrt p + co{s^{ - 1}}\sqrt {\left( {1{\text{ }} - {\text{ }}p} \right)} + co{s^{ - 1}}\sqrt {\left( {1{\text{ }} - {\text{ }}q} \right)} = \dfrac{{3\pi }}{4}\]
\[co{s^{ - 1}}\sqrt p + co{s^{ - 1}}\sqrt {\left( {1{\text{ }} - {\text{ (}}\sqrt p {)^2}} \right)} + co{s^{ - 1}}\sqrt {\left( {1{\text{ }} - {\text{ (}}\sqrt q {)^2}} \right)} = \dfrac{{3\pi }}{4}\]
\[co{s^{ - 1}}\sqrt p + {\sin ^{ - 1}}\sqrt p + co{s^{ - 1}}\sqrt {\left( {1{\text{ }} - {\text{ q}}} \right)} = \dfrac{{3\pi }}{4}\] [ Since \[{\sin ^{ - 1}}x = {\cos ^{ - 1}}\sqrt {1 - {x^2}} \]]
Using the formula \[co{s^{ - 1}}x + si{n^{ - 1}}x = \dfrac{\pi }{2}\]
\[\dfrac{\pi }{2} + co{s^{ - 1}}\sqrt {1 - q} = \dfrac{{3\pi }}{4}\]
\[co{s^{ - 1}}\sqrt {1 - q} = \dfrac{{3\pi }}{4} - \dfrac{\pi }{2}\]
\[co{s^{ - 1}}\sqrt {1 - q} = \dfrac{\pi }{4}\]
\[\sqrt {1 - q} = \cos \dfrac{\pi }{4}\]
\[\sqrt {1 - q} = \dfrac{1}{{\sqrt 2 }}\]
Squaring on both sides we get,
\[1 - q = \dfrac{1}{2}\]
\[q = 1 - \dfrac{1}{2}\]
\[q = \dfrac{1}{2}\]
Hence, the correct option is D.
Note: In this question\[p\] can be rewritten as \[{(\sqrt p )^2}\]. It is important to know formulas and properties of inverse trigonometric functions to solve these kinds of questions. It is also important to remember the trigonometric ratio table to find the values of trigonometric ratios at different angles.
Formula Used:
\[{\sin ^{ - 1}}x = {\cos ^{ - 1}}\sqrt {1 - {x^2}} \]
\[{\cos^{ - 1}}x + si{n^{ - 1}}x = \dfrac{\pi }{2}\]
Complete step by step Solution:
\[co{s^{ - 1}}\sqrt p + co{s^{ - 1}}\sqrt {\left( {1{\text{ }} - {\text{ }}p} \right)} + co{s^{ - 1}}\sqrt {\left( {1{\text{ }} - {\text{ }}q} \right)} = \dfrac{{3\pi }}{4}\]
\[co{s^{ - 1}}\sqrt p + co{s^{ - 1}}\sqrt {\left( {1{\text{ }} - {\text{ (}}\sqrt p {)^2}} \right)} + co{s^{ - 1}}\sqrt {\left( {1{\text{ }} - {\text{ (}}\sqrt q {)^2}} \right)} = \dfrac{{3\pi }}{4}\]
\[co{s^{ - 1}}\sqrt p + {\sin ^{ - 1}}\sqrt p + co{s^{ - 1}}\sqrt {\left( {1{\text{ }} - {\text{ q}}} \right)} = \dfrac{{3\pi }}{4}\] [ Since \[{\sin ^{ - 1}}x = {\cos ^{ - 1}}\sqrt {1 - {x^2}} \]]
Using the formula \[co{s^{ - 1}}x + si{n^{ - 1}}x = \dfrac{\pi }{2}\]
\[\dfrac{\pi }{2} + co{s^{ - 1}}\sqrt {1 - q} = \dfrac{{3\pi }}{4}\]
\[co{s^{ - 1}}\sqrt {1 - q} = \dfrac{{3\pi }}{4} - \dfrac{\pi }{2}\]
\[co{s^{ - 1}}\sqrt {1 - q} = \dfrac{\pi }{4}\]
\[\sqrt {1 - q} = \cos \dfrac{\pi }{4}\]
\[\sqrt {1 - q} = \dfrac{1}{{\sqrt 2 }}\]
Squaring on both sides we get,
\[1 - q = \dfrac{1}{2}\]
\[q = 1 - \dfrac{1}{2}\]
\[q = \dfrac{1}{2}\]
Hence, the correct option is D.
Note: In this question\[p\] can be rewritten as \[{(\sqrt p )^2}\]. It is important to know formulas and properties of inverse trigonometric functions to solve these kinds of questions. It is also important to remember the trigonometric ratio table to find the values of trigonometric ratios at different angles.
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