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How can I solve sinA-cosA=1 where 0 degree<=A<=90 degree. Thank you. Nazrul,. One way to approach this problem is to use sin2A = 1 - cos2A and hence write. α. Cos ox. 11- cas²α sind. Sia cosa sina. Cosa. ( cos²α-sin²α. Costa. 7-. Sin α. COJ∞ cosα-sinα cosa - Sina. (casa-sinα) (cosatsinα). Cosa - Sina cosa + sina. : ... May 21, 2018 ... 2 Answers ... Dean R. ... It seemed easier to start from the right; please see proof below. Explanation: We need the crudest cosine double angle ... Jan 3, 2023 ... cos A/(1+sin A)+(1+sin A)/cos A · =cos²A+(1+sin A)²/(cos A (1+sin A)) · = (cos² A + 1+ sin² A + 2 sin A )/(cos A(1+sin A)) · = 2(1+ sin A)/cos A ... ... sina sina cosa A—, 1 +A”' +A-. +A— +A=0 1[ iJOW-4aifi 2[ ikfi-4atfi 3[ 402] d 20 l 5 Qasina cosa cosa sina A—., .+As , .+A—+A— +A=0 1[ 73(72—402)] 2[7(72 ... - cos²α cos & + cosa sina sin) cosx sina cos✪ + sin² α sin cosa ( cosa cos € + sina ( cos x cos cos α cos (α Ө sina cos (x cos (α-8) (cos α). (b) The ... cos 2 A = sin A . tan A ( G . P ) = sin A . sin A cos A cos 3 A = sin 2 A ⇒ cos 3 A sin 2 A = 1 → ( 1 ) cos 2 A sin 2 A . Sep 16, 2002 ... sina. sinA. = sinb. sinB. = sinc. sinC. Cosine Rule sides. Cosine Rule angles cosa = cosbcosc+sinbsinccosA cosb= cosccosa+sincsinacosB cosc ... Prove that: (cosA - sinA + 1) / (cosA + sinA +1) = (1 - sinA) / cosA. Answer: For solving this questions we need to rationalize the denominator. sinA. cosA. = -2cos2A. 2. a. On the ABC triangle it is known that cosA.cosB = sinA.sinB and. cosA.cosB = sinA.sinB Define the shape of the ABC triangle and the ...