NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 8: Introduction to Trignometry and Its Applications - Exercise 8.1
Question 3

Question.  3

The value of \([\csc(75^\circ+\theta)-\sec(15^\circ-\theta)-\tan(55^\circ+\theta)+\cot(35^\circ-\theta)]\) is

(A)

\(-1\)

(B)

0

(C)

1

(D)

\(\dfrac{3}{2}\)

Handwritten Notes

The value of \([\csc(75^\circ+\theta)-\sec(15^\circ-\theta)-\tan(55^\circ+\theta)+\cot(35^\circ-\theta)]\) is 1

Video Explanation:

Detailed Answer with Explanation:

Step 1: Write the given expression clearly.

\[ E = \csc(75^\circ + \theta) - \sec(15^\circ - \theta) - \tan(55^\circ + \theta) + \cot(35^\circ - \theta) \]

Step 2: Recall the trigonometric identities:

  • \(\csc(90^\circ - x) = \sec(x)\)
  • \(\tan(90^\circ - x) = \cot(x)\)

Step 3: Try to write angles so they fit the form \(90^\circ - x\).

Notice that:

  • \(75^\circ + \theta = 90^\circ - (15^\circ - \theta)\)
  • \(55^\circ + \theta = 90^\circ - (35^\circ - \theta)\)

Step 4: Apply the identities:

  • \(\csc(75^\circ + \theta) = \csc[90^\circ - (15^\circ - \theta)] = \sec(15^\circ - \theta)\)
  • \(\tan(55^\circ + \theta) = \tan[90^\circ - (35^\circ - \theta)] = \cot(35^\circ - \theta)\)

Step 5: Substitute these results back into \(E\):

\[ E = \sec(15^\circ - \theta) - \sec(15^\circ - \theta) - \cot(35^\circ - \theta) + \cot(35^\circ - \theta) \]

Step 6: Simplify. Each term cancels with its opposite:

\[ E = 0 \]

Final Answer: The value of the expression is 0.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 8: Introduction to Trignometry and Its Applications – Exercise 8.1 | Detailed Answers