SSC CGL level Solution Set 77, Trigonometry 7 | SureSolv

SSC CGL level Solution Set 77, Trigonometry 7

77th SSC CGL level Solution Set, topic Trigonometry 7

ssc cgl level solution set 77 trigonometry 7

This is the 77th solution set for the 10 practice problem exercise for SSC CGL exam and 7th on topic Trigonometry.

If you have not taken the test yet, you should refer to the 77th SSC CGL question set and 7th on Trigonometry before going through this solution.


77th solution set- 10 problems for SSC CGL exam: 7th on Trigonometry - testing time 12 mins

Problem 1.

If $7\sin^2 \theta+3\cos^2 \theta=4$, then the value of $\tan \theta$, where $\theta$ is acute, is,

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

Solution 1: Problem analysis and execution

$3\sin^2 \theta$ out of $7\sin^2 \theta$ combines with $3\cos^2 \theta$ to produce 3 and reduces the 4 in RHS to 1, and leaving $4\sin^2 \theta$ in the LHS.

So, $4\sin^2 \theta=1$,

Or, $\sin^2 \theta=\displaystyle\frac{1}{4}$,

Or, $\sin \theta=\displaystyle\frac{1}{2}$, as $\theta$ is acute $\sin \theta$ is positive,

Or, $\theta=30^0$,

Or, $\tan \theta=\displaystyle\frac{1}{\sqrt{3}}$, as $\cos 30^0=\displaystyle\frac{\sqrt{3}}{2}$.

Answer: Option b: $\displaystyle\frac{1}{\sqrt{3}}$.

Key concepts used: Basic trigonometry concepts -- Use of $\sin^2 \theta + \cos^2 \theta=1$.

Problem 2.

If $\alpha + \beta=90^0$, then the expression $\displaystyle\frac{\tan \alpha}{\tan \beta}+ \sin^2 \alpha + \sin^2 \beta$ is equal to,

  1. $\sec^2 \beta$
  2. $\sec^2 \alpha$
  3. $\tan^2 \beta$
  4. $\tan^2 \alpha$

Solution 2: Problem analysis and execution

Using complementary trigonometric function concept, we get,

$\sin \alpha=\sin (90^0-\beta)=\cos \beta$, and similarly,

$\cos \alpha=\sin \beta$, and

$\tan \beta=\text{cot } \alpha$.

So the target expression reduces to,

$\tan^2 \alpha + \sin^2 \alpha + \cos^2 \alpha$

$=1+\tan^2 \alpha$

$=\sec^2 \alpha$.

Answer: Option b: $\sec^2 \alpha$.

Key concepts used: Complementary trigonometric functions -- Trigonometric function relations.

Problem 3.

If $0^0 \lt \theta \lt 90^0$ then the value of $\displaystyle\frac{\tan \theta-\sec \theta-1}{\tan \theta + \sec \theta +1}$ is,

  1. $\displaystyle\frac{1-\cos \theta}{\sin \theta}$
  2. $\displaystyle\frac{\sin \theta +1}{\cos \theta}$
  3. $\displaystyle\frac{1-\sin \theta}{\cos \theta}$
  4. $\displaystyle\frac{\sin \theta-1}{\cos \theta}$

Solution 3: Problem analysis and pattern identification

The similarity of the numerator and denominator expressions and presence of $sec \theta$ and $\tan \theta$ in additive (or subtractive) relation urged us to apply the powerful friendly trigonometric function pair relation,

$\sec \theta - \tan \theta=\displaystyle\frac{1}{\sec \theta + \tan \theta}$, as $\sec^2 \theta -\tan^2 \theta=1$.

Solution 3: Problem Solving execution

As expected when we replaced the $-(\sec \theta-\tan \theta)$ in the numerator by $-\displaystyle\frac{1}{\sec \theta + \tan \theta}$, we got the numerator same as denominator except for a negative sign,

$E=\displaystyle\frac{\tan \theta-\sec \theta-1}{\tan \theta + \sec \theta +1}$, target expression is assumed to be $E$,

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

$=-\displaystyle\frac{1}{\sec \theta + \tan \theta}\left[\displaystyle\frac{\sec \theta + \tan \theta +1}{\sec \theta + \tan \theta + 1}\right]$

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

We will now multiply numerator and denominator by $(\sec \theta - \tan \theta)$,

$E=\tan \theta - \sec \theta$

$=\displaystyle\frac{\sin \theta - 1}{\cos \theta}$.

Answer: Option d: $\displaystyle\frac{\sin \theta-1}{\cos \theta}$.

Key concepts used: Problem analysis -- Key pattern identification -- Friendly trigonometric function pairs concepts -- Efficient simplification.

The solution could easily be carried out mentally.

Problem 4.

If $5\sin \theta=3$, then the numerical value of $\displaystyle\frac{\sec \theta - \tan \theta}{\sec \theta + \tan \theta}$ is,

  1. $\displaystyle\frac{1}{2}$
  2. $\displaystyle\frac{1}{4}$
  3. $\displaystyle\frac{1}{3}$
  4. $\displaystyle\frac{1}{5}$

Solution 4: Problem analysis and planning for using Componendo dividendo

The target expression is just right for applying the fast and quick method of componendo dividendo. One way to proceed is to assume it equal to a dummy variable $x$, execute componendo dividendo to get on the LHS a simplified form, $\displaystyle\frac{\sec \theta}{\tan \theta}$ and on the RHS, $\displaystyle\frac{1+x}{1-x}$. Answer is just a step ahead.

Let us show how.

Solution 4: Problem solving execution

As planned we have,

$\displaystyle\frac{\sec \theta-\tan \theta}{\sec \theta + \tan \theta}=x$.

Applying three-step componendo dividendo on both sides,

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

Or, $\displaystyle\frac{1+x}{1-x}=\text{cosec } \theta=\displaystyle\frac{5}{3}$, from given expression, $5\sin \theta=3$.

Applying componendo dividendo a second time,

$x=\displaystyle\frac{\sec \theta -\tan \theta}{\sec \theta + \tan \theta}=\displaystyle\frac{5-3}{5+3}=\frac{1}{4}$

Solution 4: Faster method to reach solution

We know the paired fraction of the target signature expression of componendo dividendo in our case is, 

$\displaystyle\frac{\sec \theta}{\tan \theta}=\text{cosec } \theta=\displaystyle\frac{5}{3}$.

Subtracting 1, adding 1 and dividing the two results we have,

$\displaystyle\frac{\sec \theta -\tan \theta}{\sec \theta + \tan \theta}=\frac{1}{4}$.

Answer: Option b: $\displaystyle\frac{1}{4}$.

Key concepts used: Key pattern identification -- Componendo dividendo in reverse.

Problem 5.

If $\displaystyle\frac{\sec \theta + \tan \theta}{\sec \theta - \tan \theta}=2\displaystyle\frac{51}{79}$, the value of $\sin \theta$ is,

  1. $\displaystyle\frac{39}{72}$
  2. $\displaystyle\frac{91}{144}$
  3. $\displaystyle\frac{65}{144}$
  4. $\displaystyle\frac{35}{72}$

Solution 5: Problem analysis and execution: applying componendo dividendo and keeping mixed fraction in mixed form

The given expression is just right for applying componendo dividendo and getting value of $\text{cosec } \theta$ immediately. But on the RHS the mixed fraction is a bit imposing. To speed up, as we do usually with mixed fractions, we will keep the mixed fractions in mixed form (which is inherently an integer plus a fraction) as long as possible.

As planned we have,

$\displaystyle\frac{\sec \theta+\tan \theta}{sec \theta - tan \theta}=2\displaystyle\frac{51}{79}$,

Or, $\displaystyle\frac{\sec \theta}{\tan \theta}=\displaystyle\frac{3\displaystyle\frac{51}{79}}{1\displaystyle\frac{51}{79}}$,

Or, $\text{cosec } \theta=\displaystyle\frac{288}{130}$,

Or, $\sin \theta=\displaystyle\frac{65}{144}$.

By the dual application of componendo dividendo and mixed fraction technique, answer could be reached in minimum time.

Answer: Option c: $\displaystyle\frac{65}{144}$.

Key concepts used: Key pattern identification -- Componendo dividendo -- Mixed fraction breakup technique -- Efficient simplification.

Problem 6.

The value of $\theta$ ($0 \leq \theta \leq 90^0$) satisfying $2\sin^2 \theta=3\cos \theta$ is,

  1. $60^0$
  2. $45^0$
  3. $90^0$
  4. $30^0$

Solution 6: Problem analysis and execution

As the given expression is not readily amenable to solution by derivation and the given angles are the well known ones, we have decided to test choice values.

The answer could be reached quickly by the nature of the given expression.

With $\theta=60^0$, $\sin \theta=\displaystyle\frac{\sqrt{3}}{2}$ and $\cos \theta=\displaystyle\frac{1}{2}$ which satisfy the given expression perfectly.

Solution 6: Conventional solution by derivation

$2\sin^2 \theta=3\cos \theta$,

Or, $2(1-\cos^2\theta)=3\cos \theta$,

Or, $2\cos^2 \theta+3\cos \theta-2=0$,

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

As $\theta$ is acute, $\cos \theta$ must be positive and so,

$2\cos \theta=1$,

Or, $\cos \theta =\displaystyle\frac{1}{2}$,

Or, $\theta =60^0$.

Answer: Option a: $60^0$.

Key concepts used: Choice value test for quick solution -- Trigonometric ratio values -- Solving Quadratic equation.

Problem 7.

If $a$, $b$, $c$ are the lengths of three sides of a $\triangle ABC$. If $a$, $b$ and $c$ are related by the relation, $a^2+b^2+c^2=ab+bc+ca$, then the value of $\sin^2 \text{A}+\sin^2 \text{B}+\sin^2 \text{C}$ is,

  1. $\displaystyle\frac{3}{4}$
  2. $\displaystyle\frac{9}{4}$
  3. $\displaystyle\frac{3}{2}$
  4. $\displaystyle\frac{3\sqrt{3}}{2}$

Solution 7: Problem analysis and strategy decision

This is one of those problems that seems to be very difficult at first glance. But if you are well-acquainted with algebraic patterns you will see the solution through the curtain in twenty seconds at most.

The algebraic expression calls for multiplying both sides by 2, taking all terms on the LHS and rearranging to transform the expression to a sum of three squares equal to 0. By well-known algebraic principle, each of the square terms must then be 0 resulting in, $a=b=c$, that is, an equilateral triangle with each angle $60^0$ and each $\sin$ value as $\displaystyle\frac{\sqrt{3}}{2}$. Final result, $\displaystyle\frac{9}{4}$.

Let us show the deductions.

Solution 7: Problem solving execution

We have,

$a^2+b^2+c^2=ab+bc+ca$,

Or, $2(a^2+b^2+c^2)-2(ab+bc+ca)=0$, multiplying both sides by 2 and rearranging

Or, $(a-b)^2+(b-c)^2+(c-a)^2=0$.

By the zero sum of square terms principle, each of the square term must be zero.

So,

$(a-b)=(b-c)=(c-a)=0$,

Or, $a=b=c$, all three sides of the triangle are of equal length.

Thus,

$\angle \text{A}=\angle \text{B}=\angle \text{C}=60^0$,

Or, $\sin \text{A}=\sin \text{B}=\sin \text{C}=\displaystyle\frac{\sqrt{3}}{2}$.

Finally,

$\sin^2 \text{A}+\sin^2 \text{B}+\sin^2 \text{C}=\displaystyle\frac{9}{4}$.

Answer: Option b: $\displaystyle\frac{9}{4}$.

Key concepts used: Key pattern identification -- Zero sum of square terms -- Property of an equilateral triangle -- Algebraic patterns and methods.

Problem 8.

If $x\cos^2 30^0.\sin 60^0=\displaystyle\frac{\tan^2 45^0.\sec 60^0}{\text{cosec } 60^0}$, then the value of $x$ is,

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

Solution 8: Problem analysis and execution

Instead of using the values of the trigonometric ratios and doing the involved calculations rightaway, we decide it to be the faster way to simplify the expression using patterns and trigonometric or algebraic relations first and then use the values of trigonometric ratios.

In the given expression, $\sin 60^0$ is taken to to the denominator of the RHS to cancel out $\text{cosec } 60^0$ and $\tan 45^0$ being 1, $\tan^2 45^0$ also evaporates, leaving,

$x\cos^2 30^0 =\sec 60^0=\displaystyle\frac{1}{\cos 60^0}=2$,

Or, $x\left(\displaystyle\frac{3}{4}\right)=2$, as $\cos 30^0=\displaystyle\frac{\sqrt{3}}{2}$,

Or, $x=\displaystyle\frac{8}{3}=2\displaystyle\frac{2}{3}$.

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

Key concepts used: Basic trigonometric concepts -- Values of trigonometric ratios -- Efficient simplification.

Problem 9.

If $\sec \theta-\cos \theta = \displaystyle\frac{3}{2}$, where $\theta$ is a positive acute angle, the value of $\sec \theta$ is,

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

Solution 9: Problem analysis and pattern identification

This is a special sum where the two trigonometric functions cannot be interacted with each other trigonometrically, but then we notice, one is inverse of the other. It immediately reminds us of the powerful algebraic and general principle of interaction of inverses. We become sure of the path to the solution. 

Sum (subtractive) of inverses is the crucial pattern that we will exploit.

Solution 9: Problem solving execution

The given expression,

$sec \theta-\cos \theta=\displaystyle\frac{3}{2}$.

Squaring both sides,

$\sec^2 \theta -2+\cos^2 \theta=\displaystyle\frac{9}{4}$.

Adding 4 to both sides to convert the LHS to the square of $(\sec \theta+\cos \theta)$,

$\sec^2 \theta+2+\cos^2 \theta=\displaystyle\frac{9}{4}+4=\displaystyle\frac{25}{4}$

Or, $(\sec \theta+\cos \theta)^2=\displaystyle\frac{25}{4}$

As $\theta$ is positive acute, neither $\sec \theta$ nor $\cos \theta$ can be negative, and so,

$\sec \theta+\cos \theta=\displaystyle\frac{5}{2}$.

Adding this to the given equation and simplifying,

$\sec \theta=2$

Answer: Option a: $2$.

Key concepts used: Key pattern identification -- Principle of interaction of inverses -- Algebraic patterns and methods -- Basic trigonometry concepts -- efficient simplification.

By Principle of interaction of inverses as applied to algebraic relations,

If we know the numeric value of $\left(x + \displaystyle\frac{1}{x}\right)$, we can always derive the numeric value of $\left(x-\displaystyle\frac{1}{x}\right)$ by raising the first equation to its square, adjusting the numeric middle term value and taking the square root.

This is the simplest result of the principle. Details you can find in the article, Principle of interaction of inverses.

Problem 10.

If $1+\cos^2 \theta=3\sin \theta.\cos \theta$, then the integral value of $\text{cot } \theta$ $\left(0 \lt \theta \lt \displaystyle\frac{\pi}{2}\right)$ is equal to,

  1. $0$
  2. $3$
  3. $1$
  4. $2$

Solution 10: Problem analysis

To find the quickest path to the solution in this unbalanced expression, we paused briefly resulting in time to solve extending to a minute.

The thought process was—with $(\sin \theta.\cos \theta)$ on the RHS and $\cos^2 \theta$ in the LHS, the situation called for expanding the 1 in the LHS to $(\sin^2 \theta + \cos^2 \theta)$ and transform the LHS to $(\sin \theta - \cos \theta)^2$.

This was the key to the solution.

Solution 10: Problem solving execution

The given expression,

$1+\cos^2 \theta=3\sin \theta.\cos \theta$,

Or, $\sin^2 \theta+2\cos^2 \theta=3\sin \theta.\cos \theta$,

Or, $\sin^2 \theta-2\sin\theta.\cos\theta+\cos^2\theta=\cos\theta(\sin\theta-\cos\theta)$

Or, $(\sin \theta-\cos\theta)^2=\cos\theta(\sin \theta-\cos \theta)$.

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

Thus we have two possible values of $\text{cot } \theta$,

$\sin \theta=\cos \theta$,

Or, $\text{cot } \theta=1$.

And,

$\sin \theta=2\cos \theta$, this is also possible,

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

Desired is the integral value, that is,

$\text{cot } \theta=1$.

Answer: Option c: $1$.

Key concepts and techniques used: Problem analysis and key pattern identification -- Basic algebraic concepts concepts -- Basic trigonometry concepts.

Note: 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 elegant solutions of Trigonometric problems.


Resources on Trigonometry and related topics

You may refer to our useful resources on Trigonometry and other related topics especially algebra.

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Basic and rich concepts in Trigonometry and its applications

Basic and Rich Concepts in Trigonometry part 2, proof of compound angle functions

Trigonometry concepts part 3, maxima (or minima) of Trigonometric expressions

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