GMAT General Knowledge — practice questions

Given $\alpha+\beta=\frac{\pi}{2}$, we get $\alpha=\frac\pi2-\beta$.<br><br>Also, $\beta+\gamma=\alpha$. Substituting for $\alpha$ gives $$\beta+\gamma=\frac\pi2-\beta$$ so $$\gamma=\frac\pi2-2\beta.$$<br><br>Now $$\tan\alpha=\tan(\beta+\gamma).$$ Using the addition formula, $$\tan\alpha=\frac{\tan\beta+\tan\gamma}{1-\tan\beta\tan\gamma}.$$<br><br>From $\alpha+\beta=\frac\pi2$, we have $$\tan\alpha\tan\beta=1.$$ Let $x=\tan\beta$ and $y=\tan\gamma$. Then $$\frac1x=\frac{x+y}{1-xy}.$$ This gives $$1-xy=x^2+xy$$ and hence $$1=x^2+2xy.$$ Dividing by $x$ gives $$\frac1x=x+2y.$$ Therefore $$\tan\alpha=\tan\beta+2\tan\gamma.$$<br><br>This matches option $\text{C}$.

Using the cosine rule, $$\cos A=\frac{b^2+c^2-a^2}{2bc}$$ and $$\cos B=\frac{a^2+c^2-b^2}{2ac}.$$ Now use $$\cos 2A=2\cos^2 A-1$$ and $$\cos 2B=2\cos^2 B-1.$$ Then $$b^2\cos 2A-a^2\cos 2B=2b^2\cos^2 A-2a^2\cos^2 B-(b^2-a^2).$$ Substituting the values of the cosines, $$2b^2\cos^2 A=\frac{(b^2+c^2-a^2)^2}{2c^2}$$ and $$2a^2\cos^2 B=\frac{(a^2+c^2-b^2)^2}{2c^2}.$$ Therefore, $$b^2\cos 2A-a^2\cos 2B=\frac{(b^2+c^2-a^2)^2-(a^2+c^2-b^2)^2}{2c^2}-(b^2-a^2).$$ Using difference of squares, $$\frac{[(b^2+c^2-a^2)-(a^2+c^2-b^2)][(b^2+c^2-a^2)+(a^2+c^2-b^2)]}{2c^2}-(b^2-a^2).$$ This becomes $$\frac{2(b^2-a^2)\cdot 2c^2}{2c^2}-(b^2-a^2)=2(b^2-a^2)-(b^2-a^2).$$ Hence $$b^2\cos 2A-a^2\cos 2B=b^2-a^2.$$ Comparing with the options, the correct choice is $A$.

Let $A=\sin^{-1}\left(\frac{5}{x}\right)$ and $B=\sin^{-1}\left(\frac{12}{x}\right)$. Given $A+B=\frac{\pi}{2}$, so $B=\frac{\pi}{2}-A$. Hence $\sin B=\cos A$. So $$\frac{12}{x}=\sqrt{1-\frac{25}{x^2}}.$$ Squaring, $$\frac{144}{x^2}=1-\frac{25}{x^2}.$$ Thus $$\frac{169}{x^2}=1.$$ Therefore $$x^2=169.$$ Since $\frac{5}{x}$ and $\frac{12}{x}$ must both lie in the domain of inverse sine and their inverse sines add to $\frac{\pi}{2}$, we take the positive value, so $$x=13.$$ On checking the options, this matches option $\text{C}$.