The value of log _{10} tan 1^{circ} + log _{1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Using the property $\log a + \log b = \log(ab)$,the expression becomes: $\log _{10} (	an 1^{\circ} \cdot 	an 2^{\circ} \cdot \dots \cdot 	an 89

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Using the property $\log a + \log b = \log(ab)$,the expression becomes: $\log _{10} (	an 1^{\circ} \cdot 	an 2^{\circ} \cdot \dots \cdot 	an 89^{\circ})$ We know that $	an 	heta \cdot 	an(90^{\circ} - 	heta) = 	an 	heta \cdot \cot 	heta = 1$. Pairing the terms: $(	an 1^{\circ} \cdot 	an 89^{\circ}) \cdot (	an 2^{\circ} \cdot 	an 88^{\circ}) \cdot \dots \cdot (	an 44^{\circ} \cdot 	an 46^{\circ}) \cdot 	an 45^{\circ}$ $= (1) \cdot (1) \cdot \dots \cdot (1) \cdot 1 = 1$ Therefore,$\log _{10} (1) = 0$.

The value of $\log _{10} \tan 1^{\circ} + \log _{10} \tan 2^{\circ} + \dots + \log _{10} \tan 89^{\circ}$ is equal to :-

Similar Questions

Prove that $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$.

If $\frac{\cos x}{a} = \frac{\cos (x + \theta)}{b} = \frac{\cos (x + 2\theta)}{c} = \frac{\cos (x + 3\theta)}{d}$,then $\left( \frac{a + c}{b + d} \right)$ is equal to:

$\frac{\sqrt{2} \cos 45^{\circ}+\cos 56^{\circ}+\cos 58^{\circ}-\cos 66^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}} = ?$

If $P = \tan 15^{\circ} + \cot 15^{\circ}$,$Q = \tan 22 \frac{1}{2}^{\circ} + \cot 22 \frac{1}{2}^{\circ}$ and $R = \sin 54^{\circ} + \sin 18^{\circ}$,then their ascending order is

Let $n$ be a positive integer such that $\sin \frac{\pi }{2^n} + \cos \frac{\pi }{2^n} = \frac{\sqrt{n}}{2}.$ Then

Vedclass Test Series

Exam Paper Generator

Online Exam Module