During a math class, Ms. Isha wrote the expression below on the board and asked the students to simplify it. (θ)/(1 - θ) + (1 - θ)/(θ) Jyoti solved it in her note book as follows: = (²θ + (1 - θ)²)/((1 - θ)×θ) .... (step 1) = (²θ + ²θ)/((1 - θ)×θ) .... (step 2) = (2²θ)/((1 - θ)×θ) .... (step 3) = (2θ)/((1 - θ)) .... (step 4) Identify the step in which Jyoti has made error(s), if any. Rectify the same and find the correct answer.

Error in step 2; correct answer 2θ

MathsCBSE Class 10 · Standard · Sample paper, session 2026-27 · 3 marks · Short answer

CBSE Class 10 Mathematics Standard sample paper 2026-27: question 31(A)

From the official CBSE sample question paper for 2026-27, section C, question 31(A), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q31(A) · 3 marks · Short answerIntroduction to Trigonometry

During a math class, Ms. Isha wrote the expression below on the board and asked the students to simplify it.
$\frac{c o s θ}{1 - s i n θ} + \frac{1 - s i n θ}{c o s θ}$
Jyoti solved it in her note book as follows:
$= \frac{c o s ^{2} θ + ( 1 - s i n θ ) ^{2}}{( 1 - s i n θ ) \times c o s θ}$ .... (step 1)
$= \frac{c o s ^{2} θ + c o s ^{2} θ}{( 1 - s i n θ ) \times c o s θ}$ .... (step 2)
$= \frac{2 c o s ^{2} θ}{( 1 - s i n θ ) \times c o s θ}$ .... (step 3)
$= \frac{2 c o s θ}{( 1 - s i n θ )}$ .... (step 4)
Identify the step in which Jyoti has made error(s), if any. Rectify the same and find the correct answer.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

Check step 2 carefully: is $(1 - sin θ)^{2}$ really equal to $cos^{2} θ$ ?

Hint 2

Expand $(1 - sin θ)^{2}$ , use $sin^{2} θ + cos^{2} θ = 1$ , then take out the common factor 2 and cancel $(1 - sin θ)$ .

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.

Jyoti's error is in step 2. She replaced $(1 - sin θ)^{2}$ by $cos^{2} θ$ . But $cos^{2} θ = 1 - sin^{2} θ$ , which is not the same as $(1 - sin θ)^{2}$ .
1 mark
Correct working:
$\frac{cos θ}{1 - sin θ} + \frac{1 - sin θ}{cos θ} = \frac{cos ^{2} θ + ( 1 - sin θ ) ^{2}}{( 1 - sin θ ) cos θ}$
$= \frac{cos ^{2} θ + 1 - 2 sin θ + sin ^{2} θ}{( 1 - sin θ ) cos θ}$
1 mark
Using $sin^{2} θ + cos^{2} θ = 1$ :
$= \frac{2 - 2 sin θ}{( 1 - sin θ ) cos θ}$
½ mark
$= \frac{2 ( 1 - sin θ )}{( 1 - sin θ ) cos θ} = \frac{2}{cos θ} = 2 sec θ$ .
½ mark

Answer: Error in step 2; correct answer $2 sec θ$

Where the marks are

Identifies step 2 as the wrong step (1 mark)
Expands $(1 - sin θ)^{2}$ correctly in the numerator (1 mark)
Uses $sin^{2} θ + cos^{2} θ = 1$ to get $2 - 2 sin θ$ (½ mark)
Cancels and gets $2 sec θ$ (½ mark)

Revise the chapter

Chapter: Introduction to Trigonometry (notes, formulas, where marks are lost and every past question)
Topic: Proving Simple Trigonometric Identities

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 paperSample paper 2026-27 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2026-27: question 31(A)

Hints

Step-by-step solution

Revise the chapter

More questions from Introduction to Trigonometry