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This is one of our many quizzes on Pythagorean trigonometric identity. Click on this to access more such competitive quizzes on Trigonometry.

Pythagorean trigonometric identity Quiz 2

mathematics

 Represent $\tan \theta $ in terms of $\sin \theta $ using the Pythagoras theorem of triangles.

 Assume a given ratio to some variable and represent values in the triangle and use Pythagoras theorem for finding the third side of the triangle. Then use the required ratio formula to get the answer.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5d3e7efe-10f1-438a-b652-664d47c655863472787374415121826.png"/> $ \\ \text{Consider the value of }\sin \theta = k \,\,\, \left( k<1 \right) \\ \text{Represent the above sine ratio in triangle } \\ \text{Use Pythagoras theorem} \\  \Rightarrow B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}} \\ \Rightarrow A{{C}^{2}}=1-{{k}^{2}} \\ \Rightarrow AC=\sqrt{1-{{k}^{2}}} \\  \text{Apply tangent ratio to }'\theta ' \\ \Rightarrow \tan \theta =\dfrac{AB}{AC}=\dfrac{k}{\sqrt{1-{{k}^{2}}}} \\ \Rightarrow \tan \theta =\dfrac{\sin \theta }{\sqrt{1-{{\sin }^{2}}\theta }}$

 Find  $2\cos \theta $ , if $\sin \theta =\dfrac{5}{7}$ in the first quadrant.

 Take the right angled triangle respective to the given trigonometric ratio and find the quadrant in which the given angle lies so that we get the required value.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1485f7a2-e961-43a0-9ef8-609c8161d8c63969533547876783226.png"/>$ \\ \text{Consider the value of }\sin \theta = k \,\,\, \left( k<1 \right) \\ \text{Represent the above sine ratio in triangle } \\ \text{By using the Pythagoras theorem we get the value of third side as} \\ \Rightarrow B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}} \\ \Rightarrow {{\left( 7x \right)}^{2}}={{\left( 5x \right)}^{2}}+A{{C}^{2}} \\ \Rightarrow AC=\sqrt{24{{x}^{2}}}=2x\sqrt{6} \\ \text{W.K.T in first quadrant all the trigonometric ratios are positive and formula of cosine ratio is given as} \\ \Rightarrow \cos \theta =\dfrac{AC}{BC}=\dfrac{2x\sqrt{6}}{7x} \\ \Rightarrow 2\cos \theta =2\times \dfrac{2\sqrt{6}}{7}=\dfrac{4\sqrt{6}}{7}$

 If $\cos \theta =\dfrac{-3}{5}$, find $\sin \theta $ when $\dfrac{\pi }{2}\le \theta \le \pi $.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ed6018c3-e195-4914-82a8-ebe4f2a8a71c3198698794883090008.png"/> $ \\ \text{Represent the magnitude of given trigonometric ratio in triangle } \\ \text{By using the Pythagoras theorem we get the value of third side as} \\ \Rightarrow B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}} \\ \Rightarrow {{\left( 5x \right)}^{2}}=A{{B}^{2}}+{{\left( 3x \right)}^{2}} \\ \Rightarrow AB=\sqrt{16{{x}^{2}}}=4x \\ \text{Given that the value of }\theta \text{ is in the second quadrant.} \\ \text{W.K.T in second quadrant sine trigonometric ratios is positive and formula of sine ratio is given as} \\ \Rightarrow \sin \theta =\dfrac{AB}{BC}=\dfrac{4x}{5x} \\ \Rightarrow \sin \theta =\dfrac{4}{5} $

 If  $\tan \theta =\dfrac{-3}{4}$ and $\cos \theta >0$ . Then find the sum of $\sec \theta $ in the other three quadrants.

  $\text{Given; Tangent ratio is negative and cosine ratio is positive which is possible in the fourth quadrant.} \\ \text{If }'x'\text{ is magnitude of value of }\sec \theta \text{ then sum of }\sec \theta \text{ in remaining quadrants is} \\ \Rightarrow S=x+\left( -x \right)+\left( -x \right) \\ \Rightarrow S=-x.........equation(i) \\ \text{Where, }x=\sec \theta \\ \text{Represent the modulus of given tangent ratio in triangle as } \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/84861225-5988-472a-b416-9625a9f599c92346095952334880174.png"/> $ \\ \text{By using the Pythagoras theorem we get the value of third side as} \\ \Rightarrow B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}} \\ \Rightarrow B{{C}^{2}}={{\left( 3x \right)}^{2}}+{{\left( 4x \right)}^{2}} \\ \Rightarrow BC=\sqrt{25{{x}^{2}}}=5x \\ \text{Now by using the formula of secant ratio we get} \\ \Rightarrow \sec \theta =\dfrac{BC}{AC}=\dfrac{5x}{4x} \\ \Rightarrow x=\dfrac{5}{4} \\ \text{From equation (i) we get} \\ \Rightarrow S=-x=\dfrac{-5}{4}$

 Find the value of $\sin x\times \cot x$ in the third quadrant when $\tan x=\dfrac{\sqrt{5}}{8}$

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a94f3759-5537-4e0d-b890-313667618dcd3369031974779301917.png"/>$ \\ \text{Represent the given trigonometric ratio in triangle as } \\  \text{By using the Pythagoras theorem we get the value of third side as} \\ \Rightarrow B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}} \\ \Rightarrow B{{C}^{2}}={{\left( \sqrt{5}x \right)}^{2}}+{{\left( 8x \right)}^{2}} \\ \Rightarrow BC=\sqrt{69{{x}^{2}}}=x\sqrt{69} \\ \text{Now by using the formula of sine ratio we get} \\ \Rightarrow \sin \theta =\dfrac{AB}{BC}=\dfrac{\sqrt{5}x}{x\sqrt{69}} \\ \Rightarrow \sin \theta =\dfrac{\sqrt{5}}{\sqrt{69}} \\ \text{In third quadrant we know that sine ratio is negative so we get the value of sine value as} \\ \Rightarrow \sin \theta =\dfrac{-\sqrt{5}}{\sqrt{69}} \\ \text{W.K.T in third quadrant cotangent ratio is positive then we get} \\ \Rightarrow \cot \theta =\dfrac{1}{\tan \theta } \\ \Rightarrow \cot \theta =\dfrac{8}{\sqrt{5}} \\ \text{Now let us assume that required value as} \\ \Rightarrow \sin \theta \times \cot \theta =\dfrac{-\sqrt{5}}{\sqrt{69}}\times \dfrac{8}{\sqrt{5}} \\ \Rightarrow \sin \theta \times \cot \theta =-\dfrac{8}{\sqrt{69}}$