The maximum value of $\frac{1}{\text{cosec } \theta}$, ($0 \le \theta \le 90^\circ$) is :

Step 1: Rewrite the expression

We know that the reciprocal of cosecant is sine. Therefore, the given expression can be rewritten as $\sin \theta$.

Step 2: Determine the range of $\theta$

The problem states that the angle $\theta$ lies in the range from $0^\circ$ to $90^\circ$, inclusive. This is the first quadrant.

Step 3: Find the maximum value of $\sin \theta$

In the first quadrant ($0^\circ \le \theta \le 90^\circ$), the sine function starts at $0$ (for $\theta = 0^\circ$) and increases to its maximum value of $1$ (for $\theta = 90^\circ$).

More from this chapter

• **EXERCISE 8.3** **1.** Express the trigonometric ratios sin A, tan A and sec A in terms of cot A.
• Cos A / 1 + sin a + 1 + sin a / cos A is equal to 2 sec a
• write all the other trigonometric ratios of angle A in terms of sec A
• Express all the trigonometric ratios in terms of Sin A
• Express the trigonometric ratio sign a sec a tan a in terms of cot a
• Prove that √(sec² A + csc² A) = tan A + cot A
• (secθ - tanθ) = (1)/(m) --- (2) Adding (1) & (2) secθ + tanθ + secθ - tanθ = (1)/(m) + (m)/(1) 2secθ = (1 + m²)/(m) secθ = (1 + m²)/(2m) (Hence proved) 2) If as
• If tanθ + secθ = m, prove that secθ = (m² + 1)/(2m).

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