Trigonometry — RRB NTPC Quantitative Aptitude

• Given: $\sin\theta + \cos\theta = \sqrt{2}\cos\theta$. Value of $\cos\theta - \sin\theta = ?$

• Solution:
Step 1: Rearrange the given equation: $\sin\theta = \sqrt{2}\cos\theta - \cos\theta = (\sqrt{2} - 1)\cos\theta$.
Step 2: Rationalize by multiplying both sides by $(\sqrt{2} + 1)$.
Step 3: $(\sqrt{2} + 1)\sin\theta = (\sqrt{2} + 1)(\sqrt{2} - 1)\cos\theta$.
Step 4: $\sqrt{2}\sin\theta + \sin\theta = (2 - 1)\cos\theta = \cos\theta$.
Step 5: Rearranging gives $\cos\theta - \sin\theta = \sqrt{2}\sin\theta$.

• Exam Trick: If $a\sin\theta + b\cos\theta = c$, then $b\sin\theta - a\cos\theta = \pm \sqrt{a^2+b^2-c^2}$. Here $1\sin\theta + 1\cos\theta = \sqrt{2}\cos\theta$. So $1\cos\theta - 1\sin\theta = \pm\sqrt{1^2+1^2-(\sqrt{2}\cos\theta)^2} = \pm\sqrt{2-2\cos^2\theta} = \pm\sqrt{2\sin^2\theta} = \sqrt{2}\sin\theta$.

• Final Answer: $\sqrt{2}\sin\theta$ (Option 1)

• Given: $\tan\theta + \cot\theta = 5$. Value of $\tan^2\theta + \cot^2\theta = ?$

• Solution:
Step 1: Square both sides of the given equation: $(\tan\theta + \cot\theta)^2 = 5^2$.
Step 2: Expand the left side: $\tan^2\theta + \cot^2\theta + 2\tan\theta\cot\theta = 25$.
Step 3: Since $\tan\theta$ and $\cot\theta$ are reciprocal functions, $\tan\theta\cot\theta = 1$.
Step 4: Substitute this value: $\tan^2\theta + \cot^2\theta + 2(1) = 25$.
Step 5: Therefore, $\tan^2\theta + \cot^2\theta = 25 - 2 = 23$.

• Exam Trick: This perfectly mirrors the algebraic identity $x + 1/x = k \implies x^2 + 1/x^2 = k^2 - 2$. Here $x = \tan\theta$, so the answer is directly $5^2 - 2 = 25 - 2 = 23$.

• Final Answer: $23$ (Option 3)

• Given: A series of squares of sines: $\sin^2 10^\circ + \sin^2 20^\circ + ... + \sin^2 80^\circ$.

• Solution:
Step 1: Write down the terms in the series: $\sin^2 10^\circ, \sin^2 20^\circ, \sin^2 30^\circ, \sin^2 40^\circ, \sin^2 50^\circ, \sin^2 60^\circ, \sin^2 70^\circ, \sin^2 80^\circ$.
Step 2: Use the complementary angle property: if $A + B = 90^\circ$, then $\sin^2 A + \sin^2 B = 1$.
Step 3: Pair the terms whose angles sum to $90^\circ$:
$(\sin^2 10^\circ + \sin^2 80^\circ) = 1$
$(\sin^2 20^\circ + \sin^2 70^\circ) = 1$
$(\sin^2 30^\circ + \sin^2 60^\circ) = 1$
$(\sin^2 40^\circ + \sin^2 50^\circ) = 1$
Step 4: Sum these pairs: $1 + 1 + 1 + 1 = 4$.

• Exam Trick: Find the number of terms: $n = (80 - 10)/10 + 1 = 8$. Since all terms pair perfectly without a middle odd term left over, the sum is exactly half the number of terms: $8 / 2 = 4$.

• Final Answer: $4$ (Option 1)

• Given: $\sec\theta + \tan\theta = 3$. Value of $\sin\theta = ?$

• Solution:
Step 1: We know the identity $\sec^2\theta - \tan^2\theta = 1$, which factors to $(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1$.
Step 2: Since $\sec\theta + \tan\theta = 3$, it must be that $\sec\theta - \tan\theta = 1/3$.
Step 3: Add the two equations:
$\sec\theta + \tan\theta = 3$
$\sec\theta - \tan\theta = 1/3$
$2\sec\theta = 3 + 1/3 = 10/3 \implies \sec\theta = 5/3$.
Step 4: Since $\sec\theta = \text{Hypotenuse} / \text{Base} = 5/3$, we have a right triangle with Hypotenuse $H = 5$ and Base $B = 3$.
Step 5: By Pythagorean theorem, Perpendicular $P = \sqrt{5^2 - 3^2} = \sqrt{16} = 4$.
Step 6: Therefore, $\sin\theta = \text{Perpendicular} / \text{Hypotenuse} = 4/5$.

• Exam Trick: If $\sec\theta + \tan\theta = x$, then $\sin\theta = \frac{x^2-1}{x^2+1}$. Here $x=3$, so $\sin\theta = \frac{3^2-1}{3^2+1} = \frac{8}{10} = \frac{4}{5}$. Solved in 5 seconds.

• Final Answer: $4/5$ (Option 2)

• Given: Expression $\frac{\cos 15^\circ + \sin 15^\circ}{\cos 15^\circ - \sin 15^\circ}$.

• Solution:
Step 1: Divide both the numerator and the denominator by $\cos 15^\circ$.
Step 2: Numerator becomes $1 + \tan 15^\circ$.
Step 3: Denominator becomes $1 - \tan 15^\circ$.
Step 4: The expression is now $\frac{1 + \tan 15^\circ}{1 - \tan 15^\circ}$.
Step 5: Recall the formula $\tan(45^\circ + A) = \frac{\tan 45^\circ + \tan A}{1 - \tan 45^\circ \tan A} = \frac{1 + \tan A}{1 - \tan A}$.
Step 6: Substitute $A = 15^\circ$. The expression equals $\tan(45^\circ + 15^\circ) = \tan 60^\circ$.
Step 7: The value of $\tan 60^\circ$ is $\sqrt{3}$.

• Exam Trick: Recognize the pattern $\frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} = \tan(45^\circ + \theta)$. For $\theta = 15^\circ$, this is simply $\tan 60^\circ = \sqrt{3}$.

• Final Answer: $\sqrt{3}$ (Option 1)

• Given: $\tan A = 1/2$, $\tan B = 1/3$. Value of $A+B = ?$

• Solution:
Step 1: Use the addition formula for tangent: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Step 2: Substitute the given values: $\tan(A + B) = \frac{1/2 + 1/3}{1 - (1/2)(1/3)}$.
Step 3: Simplify the numerator: $1/2 + 1/3 = \frac{3+2}{6} = 5/6$.
Step 4: Simplify the denominator: $1 - 1/6 = 5/6$.
Step 5: The expression evaluates to $\frac{5/6}{5/6} = 1$.
Step 6: Since $\tan(A + B) = 1$, the angle $(A + B)$ must be $45^\circ$ (assuming acute angles).

• Exam Trick: The pair $\tan A = 1/2, \tan B = 1/3 \implies A+B=45^\circ$ is a very common standard result in competitive exams. Memorize it to save computation time.

• Final Answer: $45^\circ$ (Option 1)

• Given: $A + B = 45^\circ$. Value of $(1 + \tan A)(1 + \tan B) = ?$

• Solution:
Step 1: Take tangent on both sides of $A+B = 45^\circ$. So, $\tan(A+B) = \tan 45^\circ = 1$.
Step 2: Expand using the formula: $\frac{\tan A + \tan B}{1 - \tan A \tan B} = 1$.
Step 3: Cross-multiply: $\tan A + \tan B = 1 - \tan A \tan B$.
Step 4: Rearrange: $\tan A + \tan B + \tan A \tan B = 1$.
Step 5: Add 1 to both sides: $1 + \tan A + \tan B + \tan A \tan B = 1 + 1 = 2$.
Step 6: Factor the left side: $(1 + \tan A)(1 + \tan B) = 2$.

• Exam Trick: For value putting, let $A = 45^\circ$ and $B = 0^\circ$ (so $A+B=45^\circ$). Then $(1 + \tan 45^\circ)(1 + \tan 0^\circ) = (1 + 1)(1 + 0) = 2 imes 1 = 2$. Value putting solves this instantly.

• Final Answer: $2$ (Option 1)

• Given: $\cos^4\theta - \sin^4\theta = \frac{2}{3}$. Value of $1 - 2\sin^2\theta = ?$

• Solution:
Step 1: The given expression is a difference of squares: $(x^2 - y^2) = (x-y)(x+y)$.
Step 2: $\cos^4\theta - \sin^4\theta = (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)$.
Step 3: We know the fundamental identity $\cos^2\theta + \sin^2\theta = 1$.
Step 4: Therefore, $\cos^4\theta - \sin^4\theta = (\cos^2\theta - \sin^2\theta) \times 1 = \cos^2\theta - \sin^2\theta$.
Step 5: Substitute $\cos^2\theta = 1 - \sin^2\theta$ into this result.
Step 6: $(1 - \sin^2\theta) - \sin^2\theta = 1 - 2\sin^2\theta$.
Step 7: Since the original expression equals $2/3$, the required expression $1 - 2\sin^2\theta$ also exactly equals $2/3$.

• Exam Trick: Remember the equivalences for $\cos 2\theta$: $\cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta = \cos^4\theta - \sin^4\theta$. They all represent the same thing. Thus, the value is unchanged: $2/3$.

• Final Answer: $2/3$ (Option 1)

• Given: Expression $\frac{\sin 5x - \sin 3x}{\cos 5x + \cos 3x}$.

• Solution:
Step 1: Use the sum-to-product formula for the numerator: $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$.
Step 2: Numerator $= 2 \sin(\frac{5x-3x}{2}) \cos(\frac{5x+3x}{2}) = 2 \sin(x) \cos(4x)$.
Step 3: Use the sum-to-product formula for the denominator: $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$.
Step 4: Denominator $= 2 \cos(\frac{5x+3x}{2}) \cos(\frac{5x-3x}{2}) = 2 \cos(4x) \cos(x)$.
Step 5: Substitute back into the fraction: $\frac{2 \sin(x) \cos(4x)}{2 \cos(4x) \cos(x)}$.
Step 6: Cancel the common terms $2$ and $\cos(4x)$ to get $\frac{\sin(x)}{\cos(x)}$.
Step 7: $\frac{\sin(x)}{\cos(x)} = \tan x$.

• Exam Trick: Use value putting to verify. Let $x = 30^\circ$. Numerator: $\sin 150^\circ - \sin 90^\circ = 0.5 - 1 = -0.5$. Denominator: $\cos 150^\circ + \cos 90^\circ = -\frac{\sqrt{3}}{2} + 0 = -\frac{\sqrt{3}}{2}$. Fraction: $\frac{-0.5}{-\sqrt{3}/2} = \frac{1}{\sqrt{3}}$. Option 1: $\tan 30^\circ = 1/\sqrt{3}$. Matches perfectly.

• Final Answer: $\tan x$ (Option 1)

• Given: Product $\sin 20^\circ \sin 40^\circ \sin 80^\circ$.

• Solution:
Step 1: Notice the pattern of angles: $20^\circ$, $40^\circ$, $80^\circ$. This fits the form $\theta, (60^\circ - \theta), (60^\circ + \theta)$ where $\theta = 20^\circ$.
Step 2: $40^\circ = 60^\circ - 20^\circ$ and $80^\circ = 60^\circ + 20^\circ$.
Step 3: Use the standard multiplicative series formula: $\sin\theta \sin(60^\circ-\theta) \sin(60^\circ+\theta) = \frac{1}{4}\sin 3\theta$.
Step 4: Substitute $\theta = 20^\circ$ into the formula: $\frac{1}{4}\sin(3 \times 20^\circ) = \frac{1}{4}\sin 60^\circ$.
Step 5: The value of $\sin 60^\circ$ is $\frac{\sqrt{3}}{2}$.
Step 6: Multiply: $\frac{1}{4} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$.

• Exam Trick: Whenever you see the $20-40-80$ sequence for sine or cosine, the result involves $\frac{1}{4}$ of the sine/cosine of three times the smallest angle. For sine: $\frac{1}{4}\sin 60^\circ = \frac{\sqrt{3}}{8}$. For cosine: $\frac{1}{4}\cos 60^\circ = \frac{1}{8}$. Direct memory saves minutes.

• Final Answer: $\frac{\sqrt{3}}{8}$ (Option 1)