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Question

# If tan (π/4 + x) + tan (π/4 − x) = a, then tan2 (π/4 + x) + tan2 (π/4 − x) = (a) a2 + 1 (b) a2 + 2 (c) a2 − 2 (d) None of these

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Solution

## (c) ${a}^{2}-2$ $\text{Given:}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}+x\right)+\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}-x\right)=a\phantom{\rule{0ex}{0ex}}⇒{\left[\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}+x\right)+\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}-x\right)\right]}^{2}={a}^{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{2}\left(\frac{\mathrm{\pi }}{4}+x\right)+{\mathrm{tan}}^{2}\left(\frac{\mathrm{\pi }}{4}-x\right)+2\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}-x\right)\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}+x\right)={a}^{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{2}\left(\frac{\mathrm{\pi }}{4}+x\right)+{\mathrm{tan}}^{2}\left(\frac{\mathrm{\pi }}{4}-x\right)={a}^{2}-2\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}-x\right)\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}+x\right)\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{2}\left(\frac{\mathrm{\pi }}{4}+x\right)+{\mathrm{tan}}^{2}\left(\frac{\mathrm{\pi }}{4}-x\right)={a}^{2}-2\left[\frac{\mathrm{tan}45°-\mathrm{tan}x}{1+\mathrm{tan}45°\mathrm{tan}x}×\frac{\mathrm{tan}45°+\mathrm{tan}x}{1-\mathrm{tan}45°\mathrm{tan}x}\right]\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{2}\left(\frac{\mathrm{\pi }}{4}+x\right)+{\mathrm{tan}}^{2}\left(\frac{\mathrm{\pi }}{4}-x\right)={a}^{2}-2\left[\frac{1°-\mathrm{tan}x}{1+\mathrm{tan}x}×\frac{1+\mathrm{tan}x}{1-\mathrm{tan}x}\right]\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{2}\left(\frac{\mathrm{\pi }}{4}+x\right)+{\mathrm{tan}}^{2}\left(\frac{\mathrm{\pi }}{4}-x\right)={a}^{2}-2\left(\frac{1-{\mathrm{tan}}^{2}x}{1-{\mathrm{tan}}^{2}x}\right)\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{2}\left(\frac{\mathrm{\pi }}{4}+x\right)+{\mathrm{tan}}^{2}\left(\frac{\mathrm{\pi }}{4}-x\right)={a}^{2}-2$

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