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

Question 6

Step 1: Write the given product:

\(P = \tan1^\circ \times \tan2^\circ \times \tan3^\circ \times \cdots \times \tan89^\circ\)

Step 2: Recall the trigonometric identity:

\(\tan(90^\circ - \theta) = \cot\theta = \dfrac{1}{\tan\theta}\)

Step 3: Pair the terms in the product. For example:

\(\tan1^\circ \times \tan89^\circ = \tan1^\circ \times \tan(90^\circ - 1^\circ) = \tan1^\circ \times \cot1^\circ = 1\)

\(\tan2^\circ \times \tan88^\circ = 1\)

\(\tan3^\circ \times \tan87^\circ = 1\)

And so on…

Step 4: Continue pairing up all terms this way. Each pair gives value = 1.

Step 5: The only angle left in the middle (which cannot be paired) is:

\(\tan45^\circ = 1\)

Step 6: Multiply everything:

\(P = (1 \times 1 \times 1 \times \cdots) \times 1 = 1\)

Final Answer: The value of the product is 1.