△ABC, sinA=cosB=tanC find the value of cos^3 A+cos^2 A−cosA.


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Question Number 176514 by CrispyXYZ last updated on 20/Sep/22

$△ A B C, \sin A = \cos B = \tan C$ $find the value of \cos^{3} A + \cos^{2} A - \cos A .$
	Answered by mr W last updated on 21/Sep/22
	$sin A=cos B ⇒B=A−(π/2) C=π−A−B=((3π)/2)−2A tan C=(1/(tan 2A))=((1−tan^2 A)/(2 tan A))=sin A 1−((sin^2 A)/(cos^2 A))=((2 sin^2 A)/(cos A)) ((2 cos^2 A−1)/(cos A))=2(1−cos^2 A) ⇒cos^3 A+cos^2 A−cos A=(1/2) ✓ cos A=(1/3)[4 sin (−(1/3)sin^(−1) (5/(32)) +((2kπ)/3))−1] since A>(π/2) and −1<cos A<0, the only solution is cos A=−(1/3)[4 sin ((1/3)sin^(−1) (5/(32)) )+1] ⇒A=π−cos^(−1) {(1/3)[4 sin ((1/3)sin^(−1) (5/(32)) )+1]} ≈113.767 843 060 258° A≈113.767843° ⇒sin A≈0.915186 B≈23.767843° ⇒cos B≈0.915186 C≈42.464314° ⇒tan C≈0.915186$
	$\sin A = \cos B \Rightarrow B = A - \frac{π}{2}$ $C = π - A - B = \frac{3 π}{2} - 2 A$ $\tan C = \frac{1}{\tan 2 A} = \frac{1 - \tan^{2} A}{2 \tan A} = \sin A$ $1 - \frac{\sin^{2} A}{\cos^{2} A} = \frac{2 \sin^{2} A}{\cos A}$ $\frac{2 \cos^{2} A - 1}{\cos A} = 2 (1 - \cos^{2} A)$ $\Rightarrow \cos^{3} A + \cos^{2} A - \cos A = \frac{1}{2} ✓$ $\cos A = \frac{1}{3} [4 \sin (- \frac{1}{3} \sin^{- 1} \frac{5}{32} + \frac{2 k π}{3}) - 1]$ $s i n c e A > \frac{π}{2} a n d - 1 < \cos A < 0,$ $t h e o n l y s o l u t i o n i s$ $\cos A = - \frac{1}{3} [4 \sin (\frac{1}{3} \sin^{- 1} \frac{5}{32}) + 1]$ $\Rightarrow A = π - \cos^{- 1} {\frac{1}{3} [4 \sin (\frac{1}{3} \sin^{- 1} \frac{5}{32}) + 1]}$ $\approx 113.767 843 060 258 °$ $A \approx 113.767843 ° \Rightarrow \sin A \approx 0.915186$ $B \approx 23.767843 ° \Rightarrow \cos B \approx 0.915186$ $C \approx 42.464314 ° \Rightarrow \tan C \approx 0.915186$
	Commented by Tawa11 last updated on 22/Sep/22

	$Great sir$
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