If $\theta $ is an acute angle, find the denominator $ D $, wh... | Mathematics MCQ | Examoogle

MathematicsTrigonometry

If $\theta$ is an acute angle, find the denominator $D$ , when $(\cot \theta - \operatorname{cosec} \theta)^2 = \frac{1 - \cos \theta}{D}$

$1 + \sin \theta$

$-\sin \theta - 1$

$-1 - \cos \theta$

$1 + \cos \theta$

Correct Answer

📐 MA • Math Direct FormulaMathematicsTrigonometry

Option D

$1 + \cos \theta$

Quick Summary:

Convert all terms into sine and cosine ratios, simplify the numerator and denominator using standard trigonometric identities, and equate the resulting fraction to match the target form.

🎯

Crack Trigonometry with Examoogle SuperTeachers

Get Free Demo Classes & Live Mocks →

📐MAMath SolutionDirect Formula

📋 Given

Expression: $(\cot\theta - \csc\theta)^2 = \frac{1 - \cos\theta}{D}$ , where $\theta$ is an acute angle.

🔢 Formula Used

cot⁡θ=cos⁡θsin⁡θ,csc⁡θ=1sin⁡θ,sin⁡2θ+cos⁡2θ=1\cot\theta = \frac{\cos\theta}{\sin\theta}, \quad \csc\theta = \frac{1}{\sin\theta}, \quad \sin^2\theta + \cos^2\theta = 1cotθ=sinθcosθ​,cscθ=sinθ1​,sin2θ+cos2θ=1

⚡ Exam Hall Shortcut / Speed Trick

Convert all terms into sine and cosine ratios, simplify the numerator and denominator using standard trigonometric identities, and equate the resulting fraction to match the target form.

⚠️ Common Student Trap / Pitfall

Confusing the sign of the denominator or incorrectly expanding $(\cos\theta - 1)^2$ as $(1 - \cos\theta)^2$ without keeping track of the denominator term.

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Step 1: Express in terms of sine and cosine

Substitute $\cot\theta = \frac{\cos\theta}{\sin\theta}$ and $\csc\theta = \frac{1}{\sin\theta}$ into the given expression.

(cot⁡θ−csc⁡θ)2=(cos⁡θsin⁡θ−1sin⁡θ)2=(cos⁡θ−1sin⁡θ)2(\cot\theta - \csc\theta)^2 = \left(\frac{\cos\theta}{\sin\theta} - \frac{1}{\sin\theta}\right)^2 = \left(\frac{\cos\theta - 1}{\sin\theta}\right)^2(cotθ−cscθ)2=(sinθcosθ​−sinθ1​)2=(sinθcosθ−1​)2

Step 2: Square numerator and denominator

Apply the square to both numerator and denominator, noting that $(\cos\theta - 1)^2 = (1 - \cos\theta)^2$ .

=(cos⁡θ−1)2sin⁡2θ=(1−cos⁡θ)2sin⁡2θ=\frac{(\cos\theta - 1)^2}{\sin^2\theta} = \frac{(1 - \cos\theta)^2}{\sin^2\theta}=sin2θ(cosθ−1)2​=sin2θ(1−cosθ)2​

Step 3: Use fundamental trigonometric identity

Replace $\sin^2\theta$ with $1 - \cos^2\theta$ in the denominator using the identity $\sin^2\theta + \cos^2\theta = 1$ .

=(1−cos⁡θ)21−cos⁡2θ=\frac{(1 - \cos\theta)^2}{1 - \cos^2\theta}=1−cos2θ(1−cosθ)2​

Step 4: Factor the denominator and simplify

Factor the denominator $1 - \cos^2\theta$ as $(1 - \cos\theta)(1 + \cos\theta)$ and cancel out the common term $(1 - \cos\theta)$ .

=(1−cos⁡θ)2(1−cos⁡θ)(1+cos⁡θ)=1−cos⁡θ1+cos⁡θ=\frac{(1 - \cos\theta)^2}{(1 - \cos\theta)(1 + \cos\theta)} = \frac{1 - \cos\theta}{1 + \cos\theta}=(1−cosθ)(1+cosθ)(1−cosθ)2​=1+cosθ1−cosθ​

Step 5: Identify denominator D

Comparing the simplified expression $\frac{1 - \cos\theta}{1 + \cos\theta}$ with the given form $\frac{1 - \cos\theta}{D}$ , we get $D = 1 + \cos\theta$ .

D=1+cos⁡θD = 1 + \cos\thetaD=1+cosθ

✅

D is correct because simplifying $(\cot\theta - \csc\theta)^2$ results in $\frac{1 - \cos\theta}{1 + \cos\theta}$ , matching the form $\frac{1 - \cos\theta}{D}$ with denominator $D = 1 + \cos\theta$ .

Core Concepts Used

Click any tag to open in AI Tutor

Trigonometric Identities Reciprocal Ratios Algebraic Simplification

💡 EXAM TIP

Similar trigonometric manipulations frequently appear in height and distance problems as

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)