Without using trigonometric tables, evaluate the following:
(sin 49°cos 41°)2+(cos 41°sin 49°)2\Big(\dfrac{\text{sin 49°}}{\text{cos 41°}}\Big)^2 + \Big(\dfrac{\text{cos 41°}}{\text{sin 49°}}\Big)^2(cos 41°sin 49°)2+(sin 49°cos 41°)2
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Solving,
⇒(sin 49°cos 41°)2+(cos 41°sin 49°)2⇒(sin (90° - 41°)cos 41°)2+(cos 41°sin (90° - 41°))2As, sin (90 - θ) = cos θ⇒(cos 41°cos 41°)2+(cos 41°cos 41°)2⇒12+12⇒1+1⇒2.\Rightarrow \Big(\dfrac{\text{sin 49°}}{\text{cos 41°}}\Big)^2 + \Big(\dfrac{\text{cos 41°}}{\text{sin 49°}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{\text{sin (90° - 41°)}}{\text{cos 41°}}\Big)^2 + \Big(\dfrac{\text{cos 41°}}{\text{sin (90° - 41°)}}\Big)^2 \\[1em] \text{As, sin (90 - θ) = cos θ} \\[1em] \Rightarrow \Big(\dfrac{\text{cos 41°}}{\text{cos 41°}}\Big)^2 + \Big(\dfrac{\text{cos 41°}}{\text{cos 41°}}\Big)^2 \\[1em] \Rightarrow 1^2 + 1^2 \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.⇒(cos 41°sin 49°)2+(sin 49°cos 41°)2⇒(cos 41°sin (90° - 41°))2+(sin (90° - 41°)cos 41°)2As, sin (90 - θ) = cos θ⇒(cos 41°cos 41°)2+(cos 41°cos 41°)2⇒12+12⇒1+1⇒2.
Hence, (sin 49°cos 41°)2+(cos 41°sin 49°)2=2.\Big(\dfrac{\text{sin 49°}}{\text{cos 41°}}\Big)^2 + \Big(\dfrac{\text{cos 41°}}{\text{sin 49°}}\Big)^2 = 2.(cos 41°sin 49°)2+(sin 49°cos 41°)2=2.
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