In a \DeltaABC, twice of its circumradius is equal to side AB and \frac{2}{\left(cosec\right)^{4} A} + \frac{3}{\left(cosec\right)^{4} B} = \frac{6}{5} . If \frac{27}{\left(cosec\right)^{8} A} + \frac{8}{\left(cosec\right)^{8} B} = \frac{3^{p} + 2^{q}}{5^{r}} (p, q, r \in N) then find the value of (p + q

Engineering

Mathematics

Circumradius Inradius and exradius

Question

In a $∆$ ABC, twice of its circumradius is equal to side AB and $\frac{2}{{cosec}^{4} A} + \frac{3}{{cosec}^{4} B} = \frac{6}{5} .$

If $\frac{27}{{cosec}^{8} A} + \frac{8}{{cosec}^{8} B} = \frac{3^{p} + 2^{q}}{5^{r}}$ (p, q, r $\in$ N) then find the value of (p + q – 3r).

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2R = C ⇒ $∠$ C = 90°

2sin⁴A + 3cos⁴A = $\frac{6}{5}$

$\Rightarrow \frac{2 {(1 - \cos 2 A)}^{2}}{4} + \frac{3 {(1 + \cos 2 A)}^{2}}{4} = \frac{6}{5}$

$\Rightarrow \cos 2 A = \frac{- 1}{5}$

⇒ 27sin⁸A + 8cos⁸A

$= \frac{27 {(1 + \frac{1}{5})}^{4}}{2^{4}} + \frac{8 {(1 - \frac{1}{5})}^{4}}{2^{4}}$

$= \frac{27 \cdot 6^{4} + 8 \cdot 4^{4}}{2^{4} \cdot 5^{4}} = \frac{3^{7} + 2^{7}}{5^{4}}$

⇒ p = 7, q = 7, r = 4 ⇒ (p + q – 3r) = 2. Ans.