θ when expressed in terms of θ, is equal to: (A) (1+²θ)/(θ) (B) √(1+²θ) (C) (√(1+²θ))/(θ) (D) (√(1-²θ))/(θ)

MathsCBSE Class 10 · Standard · Board exam 2023, set 30/4/1 · 1 mark · MCQ

CBSE Class 10 Mathematics Standard 2023: question 10

Set in the CBSE Class 10 board exam of 2023, paper 30/4/1, section A, question 10, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q10 · 1 mark · MCQIntroduction to Trigonometry

$sec θ$ when expressed in terms of $cot θ$ , is equal to:

(A) $\frac{1 + c o t ^{2} θ}{c o t θ}$
(B) $1 + cot^{2} θ$
(C) $\frac{1 + c o t ^{2} θ}{c o t θ}$
(D) $\frac{1 - c o t ^{2} θ}{c o t θ}$

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 30/4/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Start from $sec^{2} θ = 1 + tan^{2} θ$ .

Hint 2

Write $tan θ = 1/ cot θ$ and take the square root.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Use $sec^{2} θ = 1 + tan^{2} θ = 1 + \frac{1}{cot ^{2} θ} = \frac{cot ^{2} θ + 1}{cot ^{2} θ}$ .
Taking the square root (for acute $θ$ ): $sec θ = \frac{1 + cot ^{2} θ}{cot θ}$ . So the answer is option (C).
1 mark

Answer: $\frac{1 + c o t ^{2} θ}{c o t θ}$ , option (C)

Where the marks are

Chooses option (C) (1 mark)

Revise the chapter

Chapter: Introduction to Trigonometry (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Introduction to Trigonometry, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 30/4/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2023: question 10

Hints

Step-by-step solution

Revise the chapter

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