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IAS Coaching in Kota

Kota is considered to be the hub of coaching classes and various esteemed UPSC coaching institutes have also been introduced... View Article

Solve for x: 4x2 + 14x + 6 = 0.

The given expression is \(\begin{array}{l}4 \mathrm{x}^{2}+14 \mathrm{x}+6 \\ =4 \mathrm{x}^{2}+12 \mathrm{x}+2 \mathrm{x}+6 \\ =4 \mathrm{x}(\mathrm{x}+3)+2(\mathrm{x}+3) \\ =(4 \mathrm{x}+2)(\mathrm{x}+3) \\ =\mathrm{x}=\frac{-1}{2},... View Article

If p + q = 5 and pq = 6, then p3 + q3 =

Using the formula, \(\begin{array}{l}(\mathrm{a}+\mathrm{b})^{3}=\mathrm{a}^{3}+\mathrm{b}^{3}+3 \mathrm{ab}(\mathrm{a}+\mathrm{b})\\ (\mathbf{p}+\mathbf{q})^{3}=\mathbf{p}^{3}+\mathbf{q}^{3}+3 \mathbf{p q}(\mathbf{p}+\mathbf{q})\\ \mathbf{p}+\mathbf{q}=\mathbf{5} \text { and } \mathbf{p q}=\mathbf{6}\\ (\mathbf{p}+\mathbf{q})^{3}=\mathbf{p}^{3}+\mathbf{q}^{3}+3 \mathbf{p q}(\mathbf{p}+\mathbf{q})\\ \Rightarrow(5)^{3}=\mathrm{p}^{3}+\mathrm{q}^{3}+(3 \times... View Article

Prove that (sec A + tan A – 1) (sec A – tan A + 1) = 2 tan A.

The expression can be written as \(\begin{array}{l}=\left(\frac{1}{\cos \mathbf{A}}+\frac{\sin \mathbf{A}}{\cos \mathbf{A}}-1\right)\left(\frac{1}{\cos \mathbf{A}}-\frac{\sin \mathbf{A}}{\cos \mathbf{A}}+1\right)\\ =\left(\frac{1+\sin A-\cos A}{\cos A}\right)\left(\frac{1-\sin A+\cos \mathbf{A}}{\cos A}\right)\\... View Article