In $\Delta PQR $, $\angle Q = 90^\circ $. If $\tan R = \frac{1... | Mathematics MCQ | Examoogle

MathematicsTrigonometry

In $\Delta PQR$ , $\angle Q = 90°$ . If $\tan R = \frac{1}{3}$ , then the value of $1 + \frac{\sec P(3\cos R - \sin P)}{\text{cosec} R (4\sin R - \cos P)}$ is:

Correct Answer

📐 MA • Math Direct SubstitutionMathematicsTrigonometry

Option D

Quick Summary:

Use a right-angled triangle with sides 1, 3, and hypotenuse $\sqrt{10}$ . Substitute $\sin P = \frac{3}{\sqrt{10}}$ , $\cos P = \frac{1}{\sqrt{10}}$ , $\sin R = \frac{1}{\sqrt{10}}$ , and $\cos R = \frac{3}{\sqrt{10}}$ directly into the expression.

🎯

Crack Trigonometry with Examoogle SuperTeachers

Get Free Demo Classes & Live Mocks →

📐MAMath SolutionDirect Substitution

📋 Given

In triangle PQR, angle Q = 90 degrees and tan R = 1/3.

🔢 Formula Used

tan⁡R=PQQR=13\tan R = \frac{PQ}{QR} = \frac{1}{3}tanR=QRPQ​=31​, sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1sin2θ+cos2θ=1

⚡ Exam Hall Shortcut / Speed Trick

⚠️ Common Student Trap / Pitfall

Students often swap the sides for $\angle P$ and $\angle R$ , confusing the opposite and adjacent sides.

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Define triangle sides

Since $\tan R = \frac{PQ}{QR} = \frac{1}{3}$ , let $PQ = 1$ and $QR = 3$ . By the Pythagorean theorem, the hypotenuse $PR = \sqrt{1^2 + 3^2} = \sqrt{10}$ .

PR=10PR = \sqrt{10}PR=10​

Determine trigonometric ratios

Find the ratios for angles $P$ and $R$ : $\sin P = \frac{QR}{PR} = \frac{3}{\sqrt{10}}$ , $\cos P = \frac{PQ}{PR} = \frac{1}{\sqrt{10}}$ , $\sin R = \frac{PQ}{PR} = \frac{1}{\sqrt{10}}$ , and $\cos R = \frac{QR}{PR} = \frac{3}{\sqrt{10}}$ . Also, $\sec P = \frac{PR}{PQ} = \sqrt{10}$ and $\csc R = \frac{PR}{PQ} = \sqrt{10}$ .

sec⁡P=10,csc⁡R=10\sec P = \sqrt{10}, \csc R = \sqrt{10}secP=10​,cscR=10​

Substitute and simplify expression

The expression is $1 + \frac{\sec P (3 \cos R - \sin P)}{\csc R (4 \sin R - \cos P)}$ . Substituting the values: $1 + \frac{\sqrt{10}(3(\frac{3}{\sqrt{10}}) - \frac{3}{\sqrt{10}})}{\sqrt{10}(4(\frac{1}{\sqrt{10}}) - \frac{1}{\sqrt{10}})} = 1 + \frac{3(\frac{3}{\sqrt{10}}) - \frac{3}{\sqrt{10}}}{\frac{4}{\sqrt{10}} - \frac{1}{\sqrt{10}}} = 1 + \frac{\frac{9-3}{\sqrt{10}}}{\frac{4-1}{\sqrt{10}}} = 1 + \frac{6}{3} = 1 + 2 = 3$ .

1+2=31 + 2 = 31+2=3

✅

D is correct because the step-by-step substitution of trigonometric values derived from the triangle sides yields a final value of 3.

Core Concepts Used

Click any tag to open in AI Tutor

Trigonometry Right-Angled Triangle Pythagorean Theorem

💡 EXAM TIP

Always draw the triangle explicitly when trigonometric ratios of different angles (like P and R) are mixed to avoid sign errors.

Was this solution helpful?

⚡ Practice on Android App · 4.8 ★

Solve 10,000+ More Questions with AI Tutor

Get instant step-by-step AI explanations, full shift Previous Year Papers (SSC GD, RRB, SSC JE), and regional rank analysis on the official Examoogle App.

Free Download • 100K+ Aspirants

Discussion (0)

Loading discussion...

Prev Next

Correct Answer

Solve 10,000+ More Questions with AI Tutor

Related Questions

Discussion (0)