\(sin^{-1}\bigg(\frac{\sqrt3}{2}\bigg)+cos^{-1}\bigg(\frac{-\sqrt3}{2}\bigg)+tan^{-1}(-1)\)
= \(\frac{π}{3}+\frac{5π}{6}-\frac{π}{4}\)
= \(\frac{4π+10π-3π}{12}\)
=\(\frac{11π}{12}\)
Hence, the correct option is (A): \(\frac{11π}{12}\)
The direction cosines of two lines are connected by the relations \( 1 + m - n = 0 \) and \( lm - 2mn + nl = 0 \). If \( \theta \) is the acute angle between those lines, then \( \cos \theta = \) ?
Consider the following cell: $ \text{Pt}(s) \, \text{H}_2 (1 \, \text{atm}) | \text{H}^+ (1 \, \text{M}) | \text{Cr}_2\text{O}_7^{2-}, \, \text{Cr}^{3+} | \text{H}^+ (1 \, \text{M}) | \text{Pt}(s) $
Given: $ E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \, \text{V}, \quad \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
At equilibrium: $ \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
Objective: $ \text{Determine the pH at the cathode where } E_{\text{cell}} = 0. $
Various trigonometric identities are as follows:
Cosecant and Secant are even functions, all the others are odd.
T-Ratios of (2x)
sin2x = 2sin x cos x
cos 2x = cos2x – sin2x
= 2cos2x – 1
= 1 – 2sin2x
T-Ratios of (3x)
sin 3x = 3sinx – 4sin3x
cos 3x = 4cos3x – 3cosx