Question

If secA=5/4, verify that 3sinA-4sin 3 A/4cos 3 A - 3cosA = 3tanA-tan 3 A/ 1-3tan 2 A

Answer 1

ok from sec = 5/4 we can find hypo = 5k, adj side  = 4k and opp side = 3k where k is any positive no
now substituting these values in LHS and RHS we get RHS = LHS = 117/44

Answer 2

Answer:

$If \: secA=\frac{5}{4}\: then$

$\frac{3sinA-4sin^{3}A}{4cos^{3}A-3cosA}</p><p>=\frac{3tanA-tan^{3}A}{1-3tan^{2}A}$

Step-by-step explanation:

Given $secA=\frac{5}{4}$---(1)

$cosA = \frac{1}{secA}\\=\frac{4}{5}$---(2)

Now ,

$tanA = \sqrt{sec^{2}A-1}$

=$\sqrt{\left(\frac{5}{4}\right)^{2}-1}$

=$\sqrt{\frac{25}{16}-1}$

=$\sqrt\frac{(25-16)}{16}}$

=$\sqrt{\frac{9}{16}}$

=$\frac{3}{4}$ ---(3)

$LHS = \frac{3sinA-4sin^{3}A}{4cos^{3}A-3cosA}$

=$\frac{sinA(3-4sin^{2}A)}{cosA(4cos^{2}A-3)}$

= $\frac{tanA[3-4(1-cos^{2}A]}{4cos^{2}A-3}$

=$\frac{tanA(4cos^{2}A-1)}{(4cos^{2}A-3)}$

=$\frac{\frac{3}{4}\big(4\times \left(\frac{4}{5}\right)^{2}-1\big)}{4\left(\frac{4}{5}\right)^{2}-3}$

After simplification, we get

= $\frac{-117}{44}$----(4)

$RHS = \frac{3tanA-tan^{3}A}{1-3tan^{2}A}$

=$\frac{3\times\frac{3}{4}-\left(\frac{3}{4}\right)^{3}}{1-3\left(\frac{3}{4}\right)^{2}}$

= $\frac{(\frac{9}{4}-\frac{27}{64})}{1-\frac{27}{16}}$

= $\frac{\frac{144-27}{64}}{\frac{(16-27)}{16}}$

=$\frac{\frac{117}{64}}{\frac{-11}{16}}$

= $\frac{117}{(-11)(4)}$

= $\frac{-117}{44}$ --(5)

Therefore,

From (4) & (5),

$LHS = RHS$

$\frac{3sinA-4sin^{3}A}{4cos^{3}A-3cosA}</p><p>=\frac{3tanA-tan^{3}A}{1-3tan^{2}A}$

•••♪