Question

Solve them please, I have an exam tomorrow

Answer 1

Solution :

$\boxed{\bold{1.}}$

Now, L.H.S.

$=\sqrt{\frac{cotA-cosA}{cotA+cosA}}$

$=\sqrt{\frac{(\frac{1}{tanA}-\frac{1}{secA})}{(\frac{1}{tanA}+\frac{1}{secA})}}$

$=\sqrt{\frac{(\frac{secA-tanA}{secA\:tanA})}{(\frac{secA+tanA}{secA\:tanA})}}$

$=\sqrt{\frac{secA-tanA}{secA+tanA}}$

$=\sqrt{\frac{secA-tanA}{\frac{1}{secA-tanA}}}$

{ since, $sec^{2}A-tan^{2}A=1$

$\to (secA+tanA)(secA-tanA)=1$

$\to secA+tanA=\frac{1}{secA-tanA}$ }

$=\sqrt{(secA-tanA)^{2}}$

$=secA-tanA$

{ since, $\sqrt{a^{2}}=a$ }

= R.H.S.

$\to \boxed{\sqrt{\frac{cotA-cosA}{cotA+cosA}}=secA-tanA}$

Hence, proved.

$\boxed{\bold{2.}}$

Given that,

$a^{4}=113-\frac{1}{a^{4}}$

$\to a^{4}+\frac{1}{a^{4}}=113$

$\to (a^{2}+\frac{1}{a^{2}})^{2}-(2 \times a \times \frac{1}{a})=113$

$\to (a^{2}+\frac{1}{a^{2}})^{2}=115$

$\to a^{2}+\frac{1}{a^{2}}=\sqrt{115}$

$\to (a-\frac{1}{a})^{2}+(2\times a \times \frac{1}{a})=\sqrt{115}$

$\to (a-\frac{1}{a})^{2}=\sqrt{115}-2$

$\to a-\frac{1}{a}=\sqrt{\sqrt{115}-2}$

$\to a-\frac{1}{a}=3$

$where,\: \sqrt{\sqrt{115}-2}=3\:(approx.)$

$\to a^{2}-1=3a$ , since a > 0

$\implies \boxed{\bold{a^{2}-3a-1=0}}$

Hence, proved.