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Andhra Pradesh SSC (Class 10) Introduction to Trigonometry — practice questions

Explore 46 free Introduction to Trigonometry questions in Andhra Pradesh SSC (Class 10) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Introduction to Trigonometry questions and exercises with answers

Showing questions 1–30 of 46.

Which expression is equal to $\sqrt{\dfrac{1+\sin A}{1-\sin A}}$ for an acute angle $A$?Read the answer and worked solution →Choose the correct value of $\dfrac{1+\tan^2 A}{1+\cot^2 A}$.Read the answer and worked solution →Choose the correct value of $(\sec A+\tan A)(1-\sin A)$.Read the answer and worked solution →Choose the correct value of $9\sec^2 A-9\tan^2 A$.Read the answer and worked solution →Which of the following is true?Read the answer and worked solution →If $\sin A=\frac{1}{3}$ for an acute angle $A$, what is the value of $\cos A$?Read the answer and worked solution →In a right triangle $ABC$, right-angled at $B$, if $ an A=\frac{4}{3}$, then $ rac{\sin A}{\cos A}$ is equal toRead the answer and worked solution →A point on an object lies below the horizontal level of an observer. The angle formed by the line of sight with the horizontal is called theRead the answer and worked solution →In a right triangle, for an acute angle θ, sin θ is defined as:Read the answer and worked solution →In a right triangle, for an acute angle θ, cos θ is defined as:Read the answer and worked solution →In terms of sine and cosine, tan θ equals:Read the answer and worked solution →The fundamental trigonometric identity is:Read the answer and worked solution →The value of sin 30° is:Read the answer and worked solution →The value of cos 60° is:Read the answer and worked solution →The value of tan 45° is:Read the answer and worked solution →The value of sin 90° is:Read the answer and worked solution →The value of cos 0° is:Read the answer and worked solution →The identity $1 + \tan^2\theta$ equals:Read the answer and worked solution →The expression $\sin(90^\circ - \theta)$ is equal to:Read the answer and worked solution →The value of tan 0° is:Read the answer and worked solution →The reciprocal of sin θ is:Read the answer and worked solution →In a right triangle with respect to acute angle A, sin A is defined as:Read the answer and worked solution →In a right triangle, if sin A = 3/5, what is cosec A?Read the answer and worked solution →In a right triangle with hypotenuse 13 and opposite (to A) = 5, what is cos A?Read the answer and worked solution →If tan A = 4/3, find sin A.Read the answer and worked solution →In a right triangle, cos A · sec A equals:Read the answer and worked solution →Evaluate sin 30° + cos 60°.Read the answer and worked solution →Find tan 45° + cot 45°.Read the answer and worked solution →Evaluate cos 60° · sin 30° + sin 60° · cos 30°.Read the answer and worked solution →sin 67° equals which of the following (using complementary angles)?Read the answer and worked solution →

Worked practice questions

Which expression is equal to $\sqrt{\dfrac{1+\sin A}{1-\sin A}}$ for an acute angle $A$?

  1. $\sec A-\tan A$
  2. $\cosec A+\cot A$
  3. $\sec A+\tan A$
  4. $\sin A+\cos A$
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Correct answer: C. $\sec A+\tan A$

Underlying idea: rationalise the expression inside the square root using $1-\sin^2 A=\cos^2 A$. $$\sqrt{\frac{1+\sin A}{1-\sin A}}=\sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}}=\sqrt{\frac{(1+\sin A)^2}{\cos^2 A}}.$$ Since $A$ is acute, $\cos A>0$, so this becomes $$\frac{1+\sin A}{\cos A}=\frac{1}{\cos A}+\frac{\sin A}{\cos A}=\sec A+\tan A.$$ Option A is tempting because $(\sec A-\tan A)(\sec A+\tan A)=1$ so it is actually the reciprocal, not the given expression. Option B uses cosec and cot, which arise from dividing by $\sin A$, not $\cos A$.

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Choose the correct value of $\dfrac{1+\tan^2 A}{1+\cot^2 A}$.

  1. $\sec^2 A$
  2. $-1$
  3. $\cot^2 A$
  4. $\tan^2 A$
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Correct answer: D. $\tan^2 A$

Using identities, $1+\tan^2 A=\sec^2 A$ and $1+\cot^2 A=\cosec^2 A$. So $$\frac{1+\tan^2 A}{1+\cot^2 A}=\frac{\sec^2 A}{\cosec^2 A}=\frac{1/\cos^2 A}{1/\sin^2 A}=\frac{\sin^2 A}{\cos^2 A}=\tan^2 A.$$ Thus option D is correct.

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Choose the correct value of $(\sec A+\tan A)(1-\sin A)$.

  1. $\sec A$
  2. $\sin A$
  3. $\csc A$
  4. $\cos A$
Show answer and solution

Correct answer: D. $\cos A$

**Rewrite in terms of sine and cosine:** $(\frac{1}{\cos A}+\frac{\sin A}{\cos A})(1-\sin A)=\frac{1+\sin A}{\cos A}(1-\sin A)$. This becomes $\frac{1-\sin^2 A}{\cos A}=\frac{\cos^2 A}{\cos A}=\cos A$.

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