In 3rd SSC CHSL solved question set on Trigonometry, 10 selected questions with answers formed the first part followed by conceptual and quick solutions.
Following are the 10 questions in the third SSC CHSL Trigonometry Solved questions set.
Third SSC CHSL Trigonometry question set- 10 selected question - testing time 15 mins
Problem 1.
If $sin (A+B)=sin Acos B + cos Asin B$, then the value of $\sin 75^0$ is,
- $\displaystyle\frac{\sqrt{3}+1}{2}$
- $\displaystyle\frac{\sqrt{2}+1}{2\sqrt{2}}$
- $\displaystyle\frac{\sqrt{3}+1}{\sqrt{2}}$
- $\displaystyle\frac{\sqrt{3}+1}{2\sqrt{2}}$
Problem 2.
If the $A$, $B$, $C$ and $D$ are the angles of a cyclic quadrilateral, the value of $cos A + cos B + cos C + cos D$ is,
- $2$
- $0$
- $1$
- $-1$
Problem 3.
If $cos^2 \alpha - sin^2 \alpha = tan^2 \beta$, then the value of $cos^2 \beta - sin^2 \beta$ is,
- $cot^2 \beta$
- $tan^2 \alpha$
- $cot^2 \alpha$
- $tan^2 \beta$
Problem 4.
If $cos^4 \theta - sin^4 \theta=\displaystyle\frac{2}{3}$, then the value of $1 - 2sin^2 \theta$ is,
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{3}{2}$
- $0$
- $1$
Problem 5.
If $x=asec \theta$ and $y=btan \theta$, then $\displaystyle\frac{x^2}{a^2}-\displaystyle\frac{y^2}{b^2}$ is,
- $0$
- $2$
- $1$
- $-1$
Problem 6.
The value of $\displaystyle\frac{\sin \theta - 2sin^3 \theta}{2cos^3 \theta - cos \theta}$ is,
- $sin \theta$
- $cot \theta$
- $cos \theta$
- $tan \theta$
Problem 7.
If $sin \theta + cos \theta=p$ and $sec \theta + cosec \theta=q$, then the value of $q(p^2 - 1)$ is,
- $2p$
- $p$
- $2$
- $1$
Problem 8.
If $sin (3\alpha - \beta)=1$, and $\cos (2\alpha + \beta)=\displaystyle\frac{1}{2}$, then the value of $tan \alpha$ is,
- $0$
- $\displaystyle\frac{1}{\sqrt{3}}$
- $\sqrt{3}$
- $1$
Problem 9.
If $rsin \theta =\displaystyle\frac{7}{2}$ and $rcos \theta =\displaystyle\frac{7\sqrt{3}}{2}$, then the value of $r$ is,
- $3$
- $4$
- $5$
- $7$
Problem 10.
If $0^0 \lt \theta \lt 90^0$ and $2sin^2 \theta + 3cos \theta=3$, then the value of $\theta$ is,
- $30^0$
- $45^0$
- $60^0$
- $75^0$
Key concepts and techniques used: Complementary trigonometric functions -- basic trigonometry concepts.
Note: You will observe that in many of the Trigonometric problems basic and rich algebraic concepts and techniques are to be used. In fact that is the norm. Algebraic concepts are frequently used for quick solutions of Trigonometric problems.
Answers to the questions
Problem 1. Answer: Option d: $\displaystyle\frac{\sqrt{3}+1}{2\sqrt{2}}$.
Problem 2. Answer: Option b: $0$.
Problem 3. Answer: Option b: $tan^2 \alpha$.
Problem 4. Answer: Option a: $\displaystyle\frac{2}{3}$.
Problem 5. Answer: Option c: $1$.
Problem 6. Answer: Option d: $tan \theta$.
Problem 7. Answer: Option a: $2p$.
Problem 8. Answer: Option b: $\displaystyle\frac{1}{\sqrt{3}}$.
Problem 9. Answer: Option d: $7$.
Problem 10. Answer: Option c: $60^0$.
Solution to the Third SSC CHSL Trigonometry question set - testing time was 15 mins
Problem 1.
If $sin (A+B)=sin Acos B + cos Asin B$, then the value of $\sin 75^0$ is,
- $\displaystyle\frac{\sqrt{3}+1}{2}$
- $\displaystyle\frac{\sqrt{2}+1}{2\sqrt{2}}$
- $\displaystyle\frac{\sqrt{3}+1}{\sqrt{2}}$
- $\displaystyle\frac{\sqrt{3}+1}{2\sqrt{2}}$
Solution 1 - Problem analysis
Compound angle trigonometric knowledge is not wanted, rather, a compound angle relationship has been given for $sin(A+B)$.
As all the terms on the RHS are in terms of $sin A$, $sin B$, $cos A$ and $cos B$, we must select $A$ and $B$ in such a way that, $A+B=75^0$ and also the trigonometric ratio term values on the RHS are commonly used and known.
This is the main challenge for solving the problem—finding a suitable combination of $A$ and $B$.
Solution 1 - Problem solving execution
Soon as the challenge was known, we could identify the best combination as,
$A + B=45^0+30^0=75^0$.
Substituting the related values in RHS,
$sin (A+B)=\displaystyle\frac{1}{\sqrt{2}}\times{\displaystyle\frac{\sqrt{3}}{2}} + \displaystyle\frac{1}{\sqrt{2}}\times{\displaystyle\frac{1}{2}}$
$=\displaystyle\frac{\sqrt{3}+1}{2\sqrt{2}}$.
Answer: Option d: $\displaystyle\frac{\sqrt{3}+1}{2\sqrt{2}}$.
Key concepts and techniques used: Commonly used values of trigonometric ratios -- basic trigonometry concepts.
Problem 2.
If the $A$, $B$, $C$ and $D$ are the angles of a cyclic quadrilateral, the value of $cos A + cos B + cos C + cos D$ is,
- $2$
- $0$
- $1$
- $-1$
Solution 2 - Problem analysis
For a cyclic quadrilateral ABCD, opposite angles sum up to $180^0$ so that,
$cos A = cos(180^0- C)=-cos C$, and
$cos B = cos(180^0- D)=-cos D$.
So,
$cos A + cos B + cos C + cos D=0$.
We have used the relation between opposite angles of a cyclic quadrilateral and also the supplementary trigonometric function concepts.
Answer: Option b: $0$.
Key concepts and techniques used: Cyclic quadrilateral -- Supplementary trigonometric functions.
Supplementary angle trigonometric functions
Two trigonometric functions are supplementary functions if, the sum of the two angles $A$ and $C$ is $180^0$.
Some of the supplemetary angle relationships,
$cos A = cos(180^0- C)=-cos C$,
$sin A = sin(180^0- D)=sin C$,
$tan A = tan(180^0- D)=-tan C$.
Problem 3.
If $cos^2 \alpha - sin^2 \alpha = tan^2 \beta$, then the value of $cos^2 \beta - sin^2 \beta$ is,
- $cot^2 \beta$
- $tan^2 \alpha$
- $cot^2 \alpha$
- $tan^2 \beta$
Solution 3 - Problem analysis
From the target expression we find, we need to evaluate $cos^2 \beta$ and $sin^2 \beta$ in terms of functions in $\alpha$ from the given equation.
By trigonometric function derivation principle, it is always possible to transform square of any trigonometric function to square of any other trigonomtric function.
Solution 3 - Problem solving execution
Given equation,
$cos^2 \alpha - sin^2 \alpha = tan^2 \beta$
Or, $cos^2 \alpha - 1+ cos^2 \alpha = sec^2 \beta - 1$
Or, $sec^2 \beta =2cos^2 \alpha$
Or, $cos^2 \beta=\displaystyle\frac{1}{2}sec^2 \alpha$.
Or, $sin^2 \beta=1 -\displaystyle\frac{1}{2}sec^2 \alpha$.
Subtracting,
$cos^2 \beta - sin^2 \beta=sec^2 \alpha -1=tan^2 \alpha$.
Answer: Option b: $tan^2 \alpha$.
Key concepts and techniques used: Basic trigonometry concept -- Trigonometric function derivation principle.
Problem 4.
If $cos^4 \theta - sin^4 \theta=\displaystyle\frac{2}{3}$, then the value of $1 - 2sin^2 \theta$ is,
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{3}{2}$
- $0$
- $1$
Solution 4 - Problem analysis and solving
Using basic algebraic relation of $a^2 - b^2=(a+b)(a-b)$, we transfom the given equation to,
$cos^4 \theta - sin^4 \theta=\displaystyle\frac{2}{3}$
Or, $(cos^2 \theta + sin^2 \theta)(cos^2 \theta - sin^2 \theta)=\displaystyle\frac{2}{3}$
Or, $(cos^2 \theta - sin^2 \theta)=\displaystyle\frac{2}{3}$.
Or, $1 -sin^2 \theta - sin^2 \theta=\displaystyle\frac{2}{3}$,
Or, $1 -2sin^2 \theta=\displaystyle\frac{2}{3}$.
RHS remained unchanged and only LHS needed transformation.
Answer: Option a: $\displaystyle\frac{2}{3}$.
Key concepts and techniques used: Basic trigonometry concepts -- Basic algebra concepts.
Problem 5.
If $x=asec \theta$ and $y=btan \theta$, then $\displaystyle\frac{x^2}{a^2}-\displaystyle\frac{y^2}{b^2}$ is,
- $0$
- $2$
- $1$
- $-1$
Solution 5 - Problem solving execution
From the given inputs we will straightway form the ratios of $x$ and $a$ and $y$ and $b$ and take a subtraction of quares as is wanted,
$\displaystyle\frac{x^2}{a^2}-\displaystyle\frac{y^2}{b^2}=sec^2 \theta - tan^2 \theta=1$
Answer: Option c: $1$.
We used the basic relation between friendly trigonometric function pair, $sec \theta$ and $tan \theta$ and target driven algebraic simplification.
Problem 6.
The value of $\displaystyle\frac{\sin \theta - 2sin^3 \theta}{2cos^3 \theta - cos \theta}$ is,
- $sin \theta$
- $cot \theta$
- $cos \theta$
- $tan \theta$
Solution 6 - Problem analysis
$sin \theta$ and $cos \theta$ appear both in numerator and denominator in similar forms and ready to be factored out,
$\displaystyle\frac{\sin \theta - 2sin^3 \theta}{2cos^3 \theta - cos \theta}$
$=tan \theta\displaystyle\frac{1 - 2sin^2 \theta}{2cos^2 \theta - 1}$
$=tan \theta\displaystyle\frac{cos^2 \theta - sin^2 \theta}{cos^2 \theta - sin^2 \theta}$
$=tan \theta$
Answer: Option d: $tan \theta$.
Key concepts and techniques used: Basic trigonometry copncepts -- Factoring out technique -- Use of friendly trigonometric function pair, $sin \theta$ and $cos \theta$.
Problem 7.
If $sin \theta + cos \theta=p$ and $sec \theta + cosec \theta=q$, then the value of $q(p^2 - 1)$ is,
- $2p$
- $p$
- $2$
- $1$
Solution 7 - Problem solving execution
We will straightaway start deriving the value of target expression with larger factor first,
$p^2-1$,
$=(sin \theta +cos \theta)^2 - 1$
$=2sin \theta.cos \theta$, as $sin^2 \theta + cos^2 \theta=1$
So,
$q(p^2-1)$
$=(2sin \theta.cos \theta)(sec \theta + cosec \theta)$
$=2(sin \theta + cos \theta)$
$=2p$.
Answer: Option a: $2p$.
Key concepts used: Basic trigonometry concepts -- Basic algebra concepts -- Pattern identification technique -- Substitution.
Problem 8.
If $sin (3\alpha - \beta)=1$, and $\cos (2\alpha + \beta)=\displaystyle\frac{1}{2}$, then the value of $tan \alpha$ is,
- $0$
- $\displaystyle\frac{1}{\sqrt{3}}$
- $\sqrt{3}$
- $1$
Solution 8 - Problem solving execution
We will directly evaluate the bracketed angles from given trigonometric term ratio values,
$sin (3\alpha - \beta)=1$.
So,
$3\alpha - \beta=90^0$.
Again,
$cos (2\alpha + \beta)=\displaystyle\frac{1}{2}$
So,
$2\alpha + \beta=60^0$.
Adding the two,
$5 \alpha=150^0$,
Or, $\alpha=30^0$,
Or, $tan \alpha=tan 30^0 =\displaystyle\frac{1}{\sqrt{3}}$.
Answer: Option b: $\displaystyle\frac{1}{\sqrt{3}}$.
Key concepts and techniques used: Basic trigonometry concepts -- Trigonometric term values.
Problem 9.
If $rsin \theta =\displaystyle\frac{7}{2}$ and $rcos \theta =\displaystyle\frac{7\sqrt{3}}{2}$, then the value of $r$ is,
- $3$
- $4$
- $5$
- $7$
Solution 9 - Problem solving execution
We straightaway eliminate both $sin \theta$ and $cos \theta$ by first isolation (dividing by $r$), squaring and then adding,
$sin^2 \theta + cos^2 \theta = \displaystyle\frac{49}{4r^2}+\displaystyle\frac{3\times{49}}{4r^2}$
Or, $r^2=49$,
Or, $r=7$.
Answer: Option d: $7$.
Key concepts and techniques used: End state analysis approach -- Basic trigonometry concepts -- efficient simplification.
All in mind and in quick time.
Problem 10.
If $0^0 \lt \theta \lt 90^0$ and $2sin^2 \theta + 3cos \theta=3$, then the value of $\theta$ is,
- $30^0$
- $45^0$
- $60^0$
- $75^0$
Solution 10 - Problem analysis
The equation involves $cos \theta$ and $sin^2 \theta$. If only we convert $sin^2 \theta$ to $1-cos^2 \theta$, we would get a quadratic equation in $cos \theta$.
Out of the two possible roots of $cos \theta$ we expect one to be eliminated by the first range condition on $\theta$.
Solution 10. Problem solving execution
Given equation,
$2sin^2 \theta + 3cos \theta=3$
Or, $2 - 2cos^2 \theta + 3cos \theta=3$,
Or, $2cos^2 \theta - 3cos \theta + 1=0$.
Or, $(2cos \theta - 1)(cos \theta - 1)=0$.
But as, $ 0^0 \lt \theta \lt 90^0$, $cos \theta \neq 1$ leaving, $2cos \theta = 1$, or, $\theta=60^0$.
Answer: Option c: $60^0$.
Key concepts and techniques used: Basic trigonometry concepts -- basic algebraic concepts -- roots of quadratic equation -- trigonometric ratio term values.
Note: You will observe that in many of the Trigonometric problems algebraic concepts and techniques are to be used. In fact that is the norm. Algebraic concepts are frequently used for quick solutions of Trigonometric problems.