If $\theta$ is an acute angle, find the value of $D$ when $(\s... | Mathematics MCQ | Examoogle

Examoogle • User • info@examoogle.com • EE-2024-8821

Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked

Back to Practice Questions

MathematicsTrigonometry

Prev Next

If $\theta$ is an acute angle, find the value of $D$ when $(\sec \theta - \tan \theta)^2 = \frac{1 - \sin \theta}{D}$ .

1 + $\sin \theta$

1 - $\sin \theta$

1 + $\cos \theta$

1 - $\cos \theta$

Correct Answer

📐 MA • Math Direct FormulaMathematicsTrigonometry

Option A

1 + $\sin \theta$

Quick Summary:

Put $\theta = 0 ($ or a small angle like 30 degrees). For $\theta = 0, ($ sec 0 - tan $0)^2$ = (1- $0)^2$ = 1. The equation becomes 1 = (1-0)/D, so D = 1. Check which option gives 1 when $\theta = 0$ : option A gives 1+0=1.

📐MAMath SolutionDirect Formula

📋 Given

An equation involving trigonometric functions of an acute angle $\theta$ : (sec $\theta -$ tan $\theta)^2 = (1 -$ sin $\theta) / D.$

🔢 Formula Used

sec⁡θ=1cos⁡θ,tan⁡θ=sin⁡θcos⁡θ,sin⁡2θ+cos⁡2θ=1\sec \theta = \frac{1}{\cos \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \sin^2 \theta + \cos^2 \theta = 1secθ=cosθ1​,tanθ=cosθsinθ​,sin2θ+cos2θ=1

⚡ Exam Hall Shortcut / Speed Trick

⚠️ Common Student Trap / Pitfall

Many students forget to apply the Pythagorean identity $\cos^2 \theta = 1 - \sin^2 \theta$ in the denominator, leading to an incorrect simplification.

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Express in Sine and Cosine

Convert the secant and tangent functions into sine and cosine ratios.

(sec⁡θ−tan⁡θ)2=(1cos⁡θ−sin⁡θcos⁡θ)2=(1−sin⁡θcos⁡θ)2(\sec \theta - \tan \theta)^2 = \left( \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} \right)^2 = \left( \frac{1 - \sin \theta}{\cos \theta} \right)^2(secθ−tanθ)2=(cosθ1​−cosθsinθ​)2=(cosθ1−sinθ​)2

Square the expression

Expand the square on both the numerator and denominator.

(1−sin⁡θ)2cos⁡2θ\frac{(1 - \sin \theta)^2}{\cos^2 \theta}cos2θ(1−sinθ)2​

Use the identity

Substitute $\cos^2 \theta = 1 - \sin^2 \theta$ and factorize the denominator.

(1−sin⁡θ)21−sin⁡2θ=(1−sin⁡θ)2(1−sin⁡θ)(1+sin⁡θ)\frac{(1 - \sin \theta)^2}{1 - \sin^2 \theta} = \frac{(1 - \sin \theta)^2}{(1 - \sin \theta)(1 + \sin \theta)}1−sin2θ(1−sinθ)2​=(1−sinθ)(1+sinθ)(1−sinθ)2​

Simplify and equate

Cancel common factors to match the given form.

1−sin⁡θ1+sin⁡θ=1−sin⁡θD  ⟹  D=1+sin⁡θ\frac{1 - \sin \theta}{1 + \sin \theta} = \frac{1 - \sin \theta}{D} \implies D = 1 + \sin \theta1+sinθ1−sinθ​=D1−sinθ​⟹D=1+sinθ

✅

A is correct because the algebraic simplification of the given expression leads to $D = 1 + \sin \theta$ .

Core Concepts Used

Click any tag to open in AI Tutor

Trigonometric Identities Algebraic Factorization Fractional Simplification

💡 EXAM TIP

This concept of converting sec and tan to sin and cos is fundamental for solving most R.H.S. verification problems in trigonometry.

Discussion (0)

Loading discussion...

Prev Next

Correct Answer

Related Questions

Discussion (0)