Question

Show that:$\tan 48°\tan 23°\tan 42°\tan 67°=1$

Answer 1

To prove:

$\tan 48°\tan 23°\tan 42°\tan 67°=1$

$\mathrm{L}.\mathrm{H}.\mathrm{S}$ $=$$\tan 48°\tan 23°\tan 42°\tan 67°$

We can write

$$\begin{gathered} \tan 48°=\tan (90°–42°) \\ =\cot 42° \\ \tan 23°=\tan (90°–67°) \\ =\cot 67° \end{gathered}$$ $[\because \tan ({90}^0-\mathrm{\theta })=\mathrm{cot\theta }]$

$\tan 48°\times \tan 23°\times \tan 42°\times \tan 67°=\tan (90°–42°)\times \tan (90°–67°)\times \tan 42°\times \tan 67°$ $\mathbf{[}\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\mathbf{(}{\mathbf{90}}^{\mathbf{0}}\mathbf{-}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathit{c}\mathit{o}\mathit{t}\mathit{\theta }\mathbf{]}$

Substitute the values in $\mathrm{L}.\mathrm{H}.\mathrm{S}$, we get

$\mathrm{L}.\mathrm{H}.\mathrm{S}$ $=\cot 42°\times \cot 67°\times \tan 42°\times \tan 67°$

$=(\cot 42°\times \tan 42°)(\cot 67°\times \tan 67°)$

$=1\times 1$ $\left[\because cot\theta =\frac{1}{tan\theta }\Rightarrow cot\theta .tan\theta =1\right]$

$=1$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

Thus,$\tan 48°\tan 23°\tan 42°\tan 67°=1$

Hence Proved