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Question: If \[{\cos ^{ - 1}}\left( {\dfrac{3}{5}} \right) - {\sin ^{ - 1}}\left( {\dfrac{4}{5}} \right) = {\c...

If cos⁡−1(35)−sin⁡−1(45)=cos⁡−1(x){\cos ^{ - 1}}\left( {\dfrac{3}{5}} \right) - {\sin ^{ - 1}}\left( {\dfrac{4}{5}} \right) = {\cos ^{ - 1}}\left( x \right), then x=x =
A. 0
B. 1
C. -1
D. 2

Explanation

Solution

Here in this question, we have to find the exact value of a given inverse trigonometric function. For this first we have to rewrite the given equation by using a inverse trigonometric identity i.e., sin⁡−1x+cos⁡−1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2} and solve by using the Sum formula of inverse sine trigonometric ratio sin⁡−1x+sin⁡−1y=sin⁡−1(x1−y2+y1−x2){\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}\left( {x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} } \right). To further simplify by standard values of trigonometric ratios to get the required solution.

Complete step by step answer:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle; the sine, cosine, tangent, cotangent, secant, or cosecant known as trigonometric function Also called circular function and its inverse are known as inverse trigonometric functions.

Consider the given question:
cos⁡−1(35)−sin⁡−1(45)=cos⁡−1(x){\cos ^{ - 1}}\left( {\dfrac{3}{5}} \right) - {\sin ^{ - 1}}\left( {\dfrac{4}{5}} \right) = {\cos ^{ - 1}}\left( x \right) --------(1)
As we know the identity of standard trigonometry: sin⁡−1x+cos⁡−1x=π2⇒cos⁡−1x=π2−sin⁡−1x{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2} \Rightarrow {\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}x, then cos⁡−1(35){\cos ^{ - 1}}\left( {\dfrac{3}{5}} \right) can be written as cos⁡−1(35)=π2−sin⁡−1(35){\cos ^{ - 1}}\left( {\dfrac{3}{5}} \right) = \dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right), then equation (1) becomes
⇒    π2−sin⁡−1(35)−sin⁡−1(45)=cos⁡−1(x)\Rightarrow \,\,\,\,\dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right) - {\sin ^{ - 1}}\left( {\dfrac{4}{5}} \right) = {\cos ^{ - 1}}\left( x \right)
⇒    π2−(sin⁡−1(35)+sin⁡−1(45))=cos⁡−1(x)\Rightarrow \,\,\,\,\dfrac{\pi }{2} - \left( {{{\sin }^{ - 1}}\left( {\dfrac{3}{5}} \right) + {{\sin }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right) = {\cos ^{ - 1}}\left( x \right) -----(2)

Now, apply the sum formula of inverse sine trigonometric ratiosin⁡−1x+sin⁡−1y=sin⁡−1(x1−y2+y1−x2){\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}\left( {x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} } \right)
Here, x=35x = \dfrac{3}{5} and y=45y = \dfrac{4}{5}
On substituting the xx and yy values in formula, we have
⇒    π2−sin⁡−1(351−(45)2+451−(35)2)=cos⁡−1(x)\Rightarrow \,\,\,\,\dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{3}{5}\sqrt {1 - {{\left( {\dfrac{4}{5}} \right)}^2}} + \dfrac{4}{5}\sqrt {1 - {{\left( {\dfrac{3}{5}} \right)}^2}} } \right) = {\cos ^{ - 1}}\left( x \right)
⇒    π2−sin⁡−1(351−1625+451−925)=cos⁡−1(x)\Rightarrow \,\,\,\,\dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{3}{5}\sqrt {1 - \dfrac{{16}}{{25}}} + \dfrac{4}{5}\sqrt {1 - \dfrac{9}{{25}}} } \right) = {\cos ^{ - 1}}\left( x \right)
Take 25 as LCM in inside both the square roots
⇒    π2−sin⁡−1(3525−1625+4525−925)=cos⁡−1(x)\Rightarrow \,\,\,\,\dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{3}{5}\sqrt {\dfrac{{25 - 16}}{{25}}} + \dfrac{4}{5}\sqrt {\dfrac{{25 - 9}}{{25}}} } \right) = {\cos ^{ - 1}}\left( x \right)
⇒    π2−sin⁡−1(35925+451625)=cos⁡−1(x)\Rightarrow \,\,\,\,\dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{3}{5}\sqrt {\dfrac{9}{{25}}} + \dfrac{4}{5}\sqrt {\dfrac{{16}}{{25}}} } \right) = {\cos ^{ - 1}}\left( x \right)
⇒    π2−sin⁡−1(35(35)2+45(45)2)=cos⁡−1(x)\Rightarrow \,\,\,\,\dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{3}{5}\sqrt {{{\left( {\dfrac{3}{5}} \right)}^2}} + \dfrac{4}{5}\sqrt {{{\left( {\dfrac{4}{5}} \right)}^2}} } \right) = {\cos ^{ - 1}}\left( x \right)

On cancelling the square and square root, then we have
⇒    π2−sin⁡−1(35⋅35+45⋅45)=cos⁡−1(x)\Rightarrow \,\,\,\,\dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{3}{5} \cdot \dfrac{3}{5} + \dfrac{4}{5} \cdot \dfrac{4}{5}} \right) = {\cos ^{ - 1}}\left( x \right)
⇒    π2−sin⁡−1(925+1625)=cos⁡−1(x)\Rightarrow \,\,\,\,\dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{9}{{25}} + \dfrac{{16}}{{25}}} \right) = {\cos ^{ - 1}}\left( x \right)
⇒    π2−sin⁡−1(9+1625)=cos⁡−1(x)\Rightarrow \,\,\,\,\dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{{9 + 16}}{{25}}} \right) = {\cos ^{ - 1}}\left( x \right)
⇒    π2−sin⁡−1(2525)=cos⁡−1(x)\Rightarrow \,\,\,\,\dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{{25}}{{25}}} \right) = {\cos ^{ - 1}}\left( x \right)
On simplification, we get
⇒    π2−sin⁡−1(1)=cos⁡−1(x)\Rightarrow \,\,\,\,\dfrac{\pi }{2} - {\sin ^{ - 1}}\left( 1 \right) = {\cos ^{ - 1}}\left( x \right)

By the standard trigonometric table the value of sin⁡(π2)=1⇒sin⁡−1(1)=π2\sin \left( {\dfrac{\pi }{2}} \right) = 1 \Rightarrow {\sin ^{ - 1}}\left( 1 \right) = \dfrac{\pi }{2}, then on substituting we have
⇒    π2−π2=cos⁡−1(x)\Rightarrow \,\,\,\,\dfrac{\pi }{2} - \dfrac{\pi }{2} = {\cos ^{ - 1}}\left( x \right)
⇒    0=cos⁡−1(x)\Rightarrow \,\,\,\,0 = {\cos ^{ - 1}}\left( x \right)
Multiply cos function on both side, we have
⇒    cos⁡(0)=cos⁡(cos⁡−1(x))\Rightarrow \,\,\,\,\cos \left( 0 \right) = \cos \left( {{{\cos }^{ - 1}}\left( x \right)} \right)
But x x−1=1x\,{x^{ - 1}} = 1. then
⇒    cos⁡(0)=x\Rightarrow \,\,\,\,\cos \left( 0 \right) = x
By the standard trigonometric table the value of cos⁡(0)=1\cos \left( 0 \right) = 1, then we get
⇒    1=x\Rightarrow \,\,\,\,1 = x
∴    x=1\therefore \,\,\,\,x = 1
Hence, the required value x=1x = 1

Therefore, option B is the correct answer.

Note: When solving the trigonometry-based questions, we have to know the definitions and table of standard angles of all six trigonometric ratios. Remember, you should know some basic formulas of trigonometry and inverse trigonometry like identities, double and half angle formulas, sum and product formulas.