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This is the second of our many quizzes on Trigonometry overview. Click on this to access more such competitive quizzes on Trigonometry.

Trigonometry overview Quiz 2

mathematics

   Find the value of $2\sin \left( \dfrac{\pi }{2}-\dfrac{\pi }{6} \right)$

  Use $\sin (A-B)=\sin A\cos B-\cos A\sin B$. Don’t forget to multiply the answer by 2 to get the final answer.

  $ 2\sin \left( \dfrac{\pi }{2}-\dfrac{\pi }{6} \right)=2\left[ \sin \dfrac{\pi }{2}\cos \dfrac{\pi }{6}-\cos \dfrac{\pi }{2}\sin \dfrac{\pi }{6} \right] \\ = 2\sin \left(\dfrac{\pi }{3}\right)=2\cdot \dfrac{\sqrt{3}}{2}=\sqrt{3} $

 Determine the inverse value of $\text{cosec}\left( 45 \right)\sec \left( 45 \right)\cot \left( 30 \right)$?

 Convert $\text{cosec}x=\dfrac{1}{\sin x}$ and $\sec x=\dfrac{1}{\cos x}$, then substitute the values and simplify

  $ \text{cosec}\left( 45 \right)\sec \left( 45 \right)\cot \left( 30 \right) \\  =\dfrac{1}{\sin \left(45^{\circ \:}\right)}\cdot \dfrac{1}{\cos \left(45^{\circ \:}\right)}\cot \left(30^{\circ \:}\right) \\ =\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}\cdot \dfrac{\sqrt{2}}{2}} = 2\sqrt{3}$

  Simplifying  ${{\cos }^{4}}\left( x \right){{\sin }^{2}}\left( x \right)-{{\cos }^{2}}\left( x \right){{\sin }^{4}}\left( x \right)$ will give the value-

 Take the common terms outside and serve the terms inside the bracket.

 $ {{\cos }^{4}}\left( x \right){{\sin }^{2}}\left( x \right)-{{\cos }^{2}}\left( x \right){{\sin }^{4}}\left( x \right) \\  \text{Taking }{{\cos }^{2}}(x)\text{ and }{{\sin }^{2}}x\text{ as common, we have } \\ \Rightarrow {{\cos }^{2}}x{{\sin }^{2}}x\left[ {{\cos }^{2}}x-{{\sin }^{2}}x \right] \\ \Rightarrow {{\cos }^{2}}x{{\sin }^{2}}x\cos (2x)  $

 Find the value of $x$, given $\sin \left( 2x \right)-\sqrt{5}\cos x=0,\text{ when }0{}^\circ  < x < 270{}^\circ$ is

  Simplify the expression and solve for $x$ and determine $x$ for the given condition.

 $ \sin \left( 2x \right)-\sqrt{5}\cos x=0 \\  \Rightarrow 2\sin x\cos x-\sqrt{5}\cos x=0 \\  \text{Taking }\cos x\text{ as common, solve the given expression}\text{.} \\  \Rightarrow \cos x\left( 2\sin x-\sqrt{5} \right)=0 \\   \cos x=0\Rightarrow x=90{}^\circ , \\  2\sin x-\sqrt{5}=0\Rightarrow x=\dfrac{\pi }{2}+2\pi n \\  \text{It is given that, }0{}^\circ \le x \le 270{}^\circ  \\  \therefore x=90{}^\circ $

  What is the product of the hypotenuse of these two triangles if it is given that $\cot B=\dfrac{\sqrt{3}}{2}$ and $\cos D=\dfrac{3}{\sqrt{2}}$. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0ae93377-5096-431e-88d1-14643406d2d65196653039994351661.png"/>

 Find the hypotenuse using the trigonometric values given and find the product.

$ \text{In triangle ABC,}\\   \text{When }\cot B=\dfrac{\sqrt{3}}{2}=\dfrac{\text{adjacent}}{\text{perpendicular}} \\  \Rightarrow AC=2,BC=\sqrt{3} \\  \therefore \text{to find AB, use Pythagoras theorem,} \\  \Rightarrow A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}} \\  \Rightarrow A{{B}^{2}}={{2}^{2}}+{{\left( \sqrt{3} \right)}^{2}} \\  \Rightarrow AB=\sqrt{7} \\  \text{In }\Delta \text{DEF, when }\cos D=\dfrac{3}{\sqrt{2}}=\dfrac{\text{adjacent}}{\text{hypotenuse}} \\  \Rightarrow DE=\sqrt{2},DF=3 \\  \text{We have hypotenuse, }DE=\sqrt{2} \\  \text{The product of the hypotenuse of triangles ABC and DEF is,}\\  \Rightarrow AB\times DE=\sqrt{14} $