SSC CHSL Solved question Set 16, Trigonometry 2 | SureSolv

SSC CHSL Solved question Set 16, Trigonometry 2

16th SSC CHSL Solved Question Set, 2nd on Trigonometry

ssc-chsl-solved-question-set-16-trigonometry-2

This is the 16th solved question set of 10 practice problem exercise for SSC CHSL exam and the 2nd on topic Trigonometry. It contains,

  • Question set on Trigonometry for SSC CHSL to be answered in 15 minutes (10 chosen questions)
  • Answers to the questions, and
  • Detailed conceptual solutions to the questions. Emphasis is on solving the problems quickly, preferably wholly in mind.

For maximum gains, the test should be taken first, and then the solutions are to be read.

IMPORTANT: To absorb the concepts, techniques and reasoning explained in the solutions fully and apply those in solving problems on Trigonometry quickly, one must solve many problems in a systematic manner using the conceptual analytical approach.

Learning by doing is the best learning. There is no other alternative towards achieving excellence.

16th Question set - 10 problems for SSC CHSL exam: 2nd on topic Trigonometry - answering time 15 mins

Q1. If $\left(\displaystyle\frac{1}{\cos \theta}\right)-\left(\displaystyle\frac{1}{\text{cot } \theta}\right)=\displaystyle\frac{1}{P}$, then what is the value of $\cos \theta$?

  1. $\displaystyle\frac{2P}{P^2+1}$
  2. $\displaystyle\frac{P+1}{P-1}$
  3. $\displaystyle\frac{P^2+1}{2P}$
  4. $\displaystyle\frac{2(P^2+1)}{P}$

Q2. The simplified value of $(\text{sec }x\text{ sec } y+\tan x\tan y)^2-(\text{sec }x\tan y+\tan x\text{ sec } y)^2$ is,

  1. $0$
  2. $-1$
  3. $1$
  4. $\text{sec}^2 x$

Q3. If $\displaystyle\frac{2\sin \theta-\cos \theta}{\cos \theta+\sin \theta}=1$, value of $\text{cot }\theta$ is,

  1. $\displaystyle\frac{1}{3}$
  2. $\displaystyle\frac{1}{2}$
  3. $2$
  4. $3$

Q4. If $\cos \theta+\text{sec } \theta=2$, the value of $\cos^6 \theta+\text{sec}^6 \theta$ is equal to,

  1. $1$
  2. $4$
  3. $8$
  4. $2$

Q5. If $2\cos \theta-\sin \theta=\displaystyle\frac{1}{\sqrt{2}}$, $(0^0 \lt \theta \lt 90^0)$, the value of $2\sin \theta+\cos \theta$ is,

  1. $\displaystyle\frac{1}{\sqrt{2}}$
  2. $\sqrt{2}$
  3. $\displaystyle\frac{\sqrt{2}}{3}$
  4. $\displaystyle\frac{3}{\sqrt{2}}$

Q6. If $\sin \theta-\cos \theta=\displaystyle\frac{7}{13}$, and $(0^0 \lt \theta \lt 90^0)$, then the value of $\sin \theta+\cos \theta$ is,

  1. $\displaystyle\frac{17}{13}$
  2. $\displaystyle\frac{1}{17}$
  3. $\displaystyle\frac{1}{13}$
  4. $\displaystyle\frac{13}{17}$

Q7. If $\tan \left(\displaystyle\frac{\pi}{2}-\displaystyle\frac{\theta}{2}\right)=\sqrt{3}$, then the value of $\cos \theta$ is,

  1. $0$
  2. $\displaystyle\frac{1}{\sqrt{2}}$
  3. $1$
  4. $\displaystyle\frac{1}{2}$

Q8. If $\sin(3x-20^0)=\cos(3y+20^0)$, then the value of $(x+y)$ is,

  1. $45^0$
  2. $30^0$
  3. $40^0$
  4. $20^0$

Q9. If $\tan^2 \alpha=1+2\tan^2 \beta$ ($\alpha$, $\beta$ are positive acute angles), then $\sqrt{2}\cos \alpha-\cos \beta$ is equal to,

  1. $0$
  2. $1$
  3. $-1$
  4. $\sqrt{2}$

Q10.  If $A$, $B$ and $C$ be the angles of a triangle, then which of the following is an incorrect relation?

  1. $\sin \displaystyle\frac{A+B}{2}=\cos \displaystyle\frac{C}{2}$
  2. $\text{cot } \displaystyle\frac{A+B}{2}=\tan \displaystyle\frac{C}{2}$
  3. $\tan \displaystyle\frac{A+B}{2}=\text{sec } \displaystyle\frac{C}{2}$
  4. $\cos \displaystyle\frac{A+B}{2}=\sin \displaystyle\frac{C}{2}$

Answers to the questions

Q1. Answer: Option a: $\displaystyle\frac{2P}{(P^2+1)}$.

Q2. Answer: Option c: $1$.

Q3. Answer: Option b: $\displaystyle\frac{1}{2}$.

Q4. Answer: Option d: $2$.

Q5. Answer: Option d: $\displaystyle\frac{3}{\sqrt{2}}$.

Q6. Answer: Option a : $\displaystyle\frac{17}{13}$.

Q7. Answer: Option d: $\displaystyle\frac{1}{2}$.

Q8. Answer: Option b: $30^0$.

Q9. Answer: Option a: $0$.

Q10. Answer: Option c: $\tan \displaystyle\frac{A+B}{2}=\text{sec } \displaystyle\frac{C}{2}$.

16th solution set - 10 problems for SSC CHSL exam: 2nd on topic Trigonometry - answering time 15 mins

Q1. If $\left(\displaystyle\frac{1}{\cos \theta}\right)-\left(\displaystyle\frac{1}{\text{cot } \theta}\right)=\displaystyle\frac{1}{P}$, then what is the value of $\cos \theta$?

  1. $\displaystyle\frac{2P}{(P^2+1)}$
  2. $\displaystyle\frac{P+1}{P-1}$
  3. $\displaystyle\frac{(P^2+1)}{2P}$
  4. $\displaystyle\frac{2(P^2+1)}{P}$

Solution 1: Immediate solution by use of Friendly trigonometric function relation

Let us first tell you about the powerful use of friendly trigonometric mutually inverse function relationship between $\text{sec }\theta$ and $\tan \theta$.

Normally we use the relation,

$\text{sec}^2 \theta-\tan^2 \theta=1$.

Break up the LHS into two factors and move one of the factors to the denominator of the RHS,

$\text{sec }\theta-\tan \theta=\displaystyle\frac{1}{\text{sec }\theta+\tan \theta}$

This is the relation we would use to simplify the given expression immediately,

$\text{sec }\theta-\tan \theta=P$,

Or, $\text{sec }\theta+\tan \theta=\displaystyle\frac{1}{P}$.

Sum up the two,

$2\text{sec }\theta=P+\displaystyle\frac{1}{P}=\displaystyle\frac{P^2+1}{P}$,

Or, $\cos \theta=\displaystyle\frac{2P}{P^2+1}$.

Answer: Option a: $\displaystyle\frac{2P}{(P^2+1)}$.

Key concepts used: Friendly mutually inverse trigonometric relation -- Variable elimination of $\tan \theta$ between two linear equations -- Solving in mind.

By using this powerful technique, the problem can be solved immediately.

Q2. The simplified value of $(\text{sec }x\text{ sec } y+\tan x\tan y)^2-(\text{sec }x\tan y+\tan x\text{ sec } y)^2$ is,

  1. $0$
  2. $-1$
  3. $1$
  4. $\text{sec}^2 x$

Solution 2: Quick solution by collection of like terms

First identify that the middle terms of both the whole squares are, equal but of opposite signs. So they would cancel out.

Among the left-out 4 terms, collect the like terms together. Combine the two terms with common factors of $\text{sec}^2 x$ and the other two terms with common factor $\tan^2 x$.

The given equation is then simplified to,

$(\text{sec }x\text{ sec } y+\tan x\tan y)^2-(\text{sec }x\tan y+\tan x\text{ sec } y)^2$

$=\text{sec}^2x(\text{sec}^2 y-\tan^2 y)-\tan^2 x(\text{sec}^2 y-\tan^2 y)$

$=\text{sec}^2 x-\tan^2 x$, as $(\text{sec}^2 y-\tan^2 y)=1$

$=1$.

Answer: Option c: $1$.

Key concepts used: Key pattern identification -- Collection of like terms -- Solving in mind.

Q3. If $\displaystyle\frac{2\sin \theta-\cos \theta}{\cos \theta+\sin \theta}=1$, value of $\text{cot }\theta$ is,

  1. $\displaystyle\frac{1}{3}$
  2. $\displaystyle\frac{1}{2}$
  3. $2$
  4. $3$

Solution 3: Immediate solution by the direct approach of cross-multiplication and collecting like terms

Cross-multiply the given equation,

$\displaystyle\frac{2\sin \theta-\cos \theta}{\cos \theta+\sin \theta}=1$,

Or, $2\sin \theta-\cos \theta=\cos \theta+\sin \theta$,

Or, $\sin \theta=2\cos \theta$.

So,

$\text{cot }\theta=\displaystyle\frac{1}{2}$

Answer: Option b: $\displaystyle\frac{1}{2}$.

Key concepts used: Direct approach -- Cross-multiplication -- Solving in mind.

Q4. If $\cos \theta+\text{sec } \theta=2$, the value of $\cos^6 \theta+\text{sec}^6 \theta$ is equal to,

  1. $1$
  2. $4$
  3. $8$
  4. $2$

Solution 4: Immediate solution by variable reduction to a quadratic equation in only $\cos \theta$ and finding the value of $\theta$ supported by mathematical reasoning

Let us tell you the mathematical reasoning first.

If you observe the high value of 6 of the power of $\cos \theta$ and $\text{sec }\theta$, you would realize that it would take a long time or it may even be very difficult to arrive at the target expression just by direct mathematical deductive steps.

Continuing in this direction of reasoning, you may be fairly certain that it should be possible to get the actual value of $\theta$ directly from the given equation.

Proceeding with this approach, covert $\text{sec }\theta$ to $\cos \theta$ to reduce the number of variables to just 1,

$\cos \theta+\text{sec } \theta=2$,

Or, $\cos^2 \theta+1=2\cos \theta$,

Or, $(1-\cos\theta)^2=0$,

Or, $\cos \theta=\text{sec }\theta=1$.

So,

$\cos^6 \theta+\text{sec}^6 \theta=1+1=2$

Answer: Option d: $2$.

Key concepts used: Mathematical reasoning -- Variable reduction principle -- Solving in mind.

Q5. If $2\cos \theta-\sin \theta=\displaystyle\frac{1}{\sqrt{2}}$, $(0^0 \lt \theta \lt 90^0)$, the value of $2\sin \theta+\cos \theta$ is,

  1. $\displaystyle\frac{1}{\sqrt{2}}$
  2. $\sqrt{2}$
  3. $\displaystyle\frac{\sqrt{2}}{3}$
  4. $\displaystyle\frac{3}{\sqrt{2}}$

Solution 5: Identify key pattern and interchange coefficients of $\sin \theta$ and $\cos \theta$ along with sign reversal by squaring and simplifying

Identify that if you can interchange the coefficients of the two variables $\sin \theta$ and $\cos \theta$ as in the given equation along with sign reversal, you would get the target expression.

First raise the given equation to square,

$2\cos \theta-\sin \theta=\displaystyle\frac{1}{\sqrt{2}}$,

Or, $4\cos^2 \theta-4\cos \theta\sin \theta+\sin^2 \theta=\displaystyle\frac{1}{2}$.

Now convert $\cos^2 \theta$ to $\sin^2 \theta$ and $\sin^2 \theta$ to $\cos^2 \theta$,

$4(1-\sin^2\theta)-4\sin \theta\cos \theta+(1-\cos^2 \theta)=\displaystyle\frac{1}{2}$,

Or, $4\sin^2 \theta+4\sin \theta \cos \theta+\cos^2 \theta=5-\displaystyle\frac{1}{2}=\displaystyle\frac{9}{2}$,

Or, $(2\sin \theta+\cos \theta)^2=\displaystyle\frac{9}{2}$.

The negative value of square root is invalid because of acute angle $\theta$. So,

$2\sin \theta+\cos \theta=\displaystyle\frac{3}{\sqrt{2}}$.

Answer: Option d. $\displaystyle\frac{3}{\sqrt{2}}$.

Key concepts used: Key pattern identification -- Coefficient interchange of two variables -- Solving in mind.

Q6. If $\sin \theta-\cos \theta=\displaystyle\frac{7}{13}$, and $(0^0 \lt \theta \lt 90^0)$, then the value of $\sin \theta+\cos \theta$ is,

  1. $\displaystyle\frac{17}{13}$
  2. $\displaystyle\frac{1}{17}$
  3. $\displaystyle\frac{1}{13}$
  4. $\displaystyle\frac{13}{17}$

Solution 6: Quick solution by coefficient interchange and sign reversal

Though both the coefficients of $\sin \theta$ and $\cos \theta$ are 1 in this case, you would achieve sign reversal by coefficient interchange method.

First square up the equation and then replace $\sin^2 \theta$ by $\cos^2 \theta$ and $\cos^2 \theta$ by $\sin^2 \theta$.

$\sin \theta-\cos \theta=\displaystyle\frac{7}{13}$,

Or, $\sin^2 \theta-2\sin \theta\cos \theta+\cos^2 \theta=\displaystyle\frac{49}{169}$,

Or, $(1-cos^2 \theta)-2\sin \theta\cos \theta+(1-\sin^2 \theta)=\displaystyle\frac{49}{169}$,

Or, $\sin^2 \theta+2\sin \theta\cos \theta+\cos^2 \theta=2-\displaystyle\frac{49}{169}=\displaystyle\frac{289}{169}$,

Or, $(\sin \theta+\cos \theta)^2=\displaystyle\frac{289}{169}$.

The negative value of square root is invalid because of acute angle $\theta$. So,

$\sin \theta+\cos \theta=\displaystyle\frac{17}{13}$.

Answer: Option a: $\displaystyle\frac{17}{13}$.

Key concepts used: Coefficient interchange with sign reversal -- Solving in mind.

Q7. If $\tan \left(\displaystyle\frac{\pi}{2}-\displaystyle\frac{\theta}{2}\right)=\sqrt{3}$, then the value of $\cos \theta$ is,

  1. $0$
  2. $\displaystyle\frac{1}{\sqrt{2}}$
  3. $1$
  4. $\displaystyle\frac{1}{2}$

Solution 7: Quick solution by using value of trigonometric functions at specific values of the angle

The given equation is,

$\tan \left(\displaystyle\frac{\pi}{2}-\displaystyle\frac{\theta}{2}\right)=\sqrt{3}=\tan 60^0$,

Or, $\left(\displaystyle\frac{\pi}{2}-\displaystyle\frac{\theta}{2}\right)=60^0$,

Or, $\theta=180^0-120^0=60^0$.

So,

$\cos \theta=\cos 60^0=\displaystyle\frac{1}{2}$.

Answer: Option: d: $\displaystyle\frac{1}{2}$.

Key concepts used: Trigonometric function values for specific angles -- Solving in mind.

Q8. If $\sin(3x-20^0)=\cos(3y+20^0)$, then the value of $(x+y)$ is,

  1. $45^0$
  2. $30^0$
  3. $40^0$
  4. $20^0$

Solution 8: Quick solution by Complementary angle trigonometric relations

Using complementary angle trigonometric relations on the given equation,

$\sin(3x-20^0)=\cos(3y+20^0)=\sin (90^0-3y-20^0)$,

Or, $3x-20^0=70^0-3y$,

Or, $3(x+y)=90^0$,

Or, $(x+y)=30^0$.

Answer: Option b: $30^0$.

Key concepts used: Complementary angle trigonometric relations -- Solving in mind.

Q9. If $\tan^2 \alpha=1+2\tan^2 \beta$ ($\alpha$, $\beta$ are positive acute angles), then $\sqrt{2}\cos \alpha-\cos \beta$ is equal to,

  1. $0$
  2. $1$
  3. $-1$
  4. $\sqrt{2}$

Solution 9: Quick solution by converting $\tan^2 \alpha$ and $\tan^2 \beta$ to $\text{sec}^2 \alpha$ and $\text{sec}^2 \beta$ by basic trigonometric identity relation

The given equation is,

$\tan^2 \alpha=1+2\tan^2 \beta$,

Or, $\text{sec}^2 \alpha -1=1+2(\text{sec}^2 \beta-1)$,

Or, $\text{sec}^2 \alpha=2\text{sec}^2 \beta$,

Or, $\sqrt{2}\cos \alpha-\cos \beta=0$.

Answer: Option a: $0$.

Key concepts used: Basic trigonometric identity relation -- Solving in mind.

Q10. If $A$, $B$ and $C$ be the angles of a triangle, then which of the following is an incorrect relation?

  1. $\sin \displaystyle\frac{A+B}{2}=\cos \displaystyle\frac{C}{2}$
  2. $\text{cot } \displaystyle\frac{A+B}{2}=\tan \displaystyle\frac{C}{2}$
  3. $\tan \displaystyle\frac{A+B}{2}=\text{sec } \displaystyle\frac{C}{2}$
  4. $\cos \displaystyle\frac{A+B}{2}=\sin \displaystyle\frac{C}{2}$

Solution 10: Quick solution by Complementary trigonometric relations

We have to test the validity of the four choices by complementary angle trigonometric relations using the triangle property of sum of three angles equals $180^0$,

$A+B+C=\pi=180^0$.

The first option,

$LHS=\sin \displaystyle\frac{A+B}{2}=\sin \left(90^0-\displaystyle\frac{C}{2}\right)$

$=\cos \left(\displaystyle\frac{C}{2}\right)$

$=RHS$.

This is a correct relation.

For the second choice,

$LHS=\text{cot } \displaystyle\frac{A+B}{2}=\text{cot } \left(90^0-\displaystyle\frac{C}{2}\right)$

$=\tan \left(\displaystyle\frac{C}{2}\right)$

$=RHS$.

This also is a correct relation.

For the third choice,

$LHS=\tan \displaystyle\frac{A+B}{2}=\tan \left(90^0-\displaystyle\frac{C}{2}\right)$

$=\text{cot } \left(\displaystyle\frac{C}{2}\right) \neq RHS$.

This is the incorrect choice.

Answer: Option c: $\tan \displaystyle\frac{A+B}{2}=\text{sec } \displaystyle\frac{C}{2}$.

Key concepts used: Sum of angle property of a triangle -- Complementary angle trigonometric relations -- Solving in mind.

End note

Observe that, each of the problems could be quickly and cleanly solved in minimum number of mental steps using special key patterns and methods in each case.

This is the hallmark of pattern based quick problem solving:

Concept based pattern and method formation, and,

Identification of the key pattern and use of the method associated with it. Every special pattern has its own method, and not many such patterns are there.

Important is the concept based pattern identification and use of quick problem solving method.


SSC CHSL level Question and Solution sets

Work and Time, Pipes and Cisterns

SSC CHSL level Solved Question set 1 on Work time 1

SSC CHSL level Solved Question set 2 on Work time 2

Number System, HCF and LCM

SSC CHSL level Solved Question set 3 on Number system 1

SSC CHSL level Solved Question set 4 on Number system 2

SSC CHSL level Solved Question set 5 on HCF and LCM 1

SSC CHSL level Solved Question set 6 on HCF and LCM 2

Surds and Indices

SSC CHSL level Solved Question set 7 on Surds and Indices 1

SSC CHSL level Solved Question set 8 on Surds and Indices 2

Mixture or Alligation

SSC CHSL level Solved Question set 9 on Mixture or Alligation 1

SSC CHSL level Solved Question set 10 on Mixture or Alligation 2

Algebra

SSC CHSL level Solved Question set 11 on Algebra 1

SSC CHSL level Solved Question set 12 on Algebra 2

SSC CHSL level Solved Question set 13 on Algebra 3

SSC CHSL level Solved Question set 14 on Algebra 4

Trigonometry

SSC CHSL level Solved Question set 15 on Trigonometry 1

SSC CHSL level Solved Question set 16 on Trigonometry 2