Evaluate the following :
5 cos260°+4 sec230°− tan245°sin230°+ cos230°\dfrac{5\text{ cos}^2 60° + 4\text{ sec}^2 30° - \text{ tan}^2 45°}{\text{sin}^2 30° + \text{ cos}^2 30°}sin230°+ cos230°5 cos260°+4 sec230°− tan245°
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Solving,
⇒5 cos260°+4 sec230°− tan245°sin230°+ cos230°⇒5×(12)2+4×(23)2−12(12)2+(32)2⇒5×14+4×43−114+34⇒54+163−144⇒15+64−12121⇒6712.\Rightarrow \dfrac{5\text{ cos}^2 60° + 4\text{ sec}^2 30° - \text{ tan}^2 45°}{\text{sin}^2 30° + \text{ cos}^2 30°} \\[1em] \Rightarrow \dfrac{5 \times \Big(\dfrac{1}{2}\Big)^2 + 4 \times \Big(\dfrac{2}{\sqrt{3}}\Big)^2 - 1^2}{\Big(\dfrac{1}{2}\Big)^2 + \Big(\dfrac{\sqrt{3}}{2}\Big)^2} \\[1em] \Rightarrow \dfrac{5 \times \dfrac{1}{4} + 4 \times \dfrac{4}{3} - 1}{\dfrac{1}{4} + \dfrac{3}{4}} \\[1em] \Rightarrow \dfrac{\dfrac{5}{4} + \dfrac{16}{3} - 1}{\dfrac{4}{4}} \\[1em] \Rightarrow \dfrac{\dfrac{15 + 64 - 12}{12}}{1} \\[1em] \Rightarrow \dfrac{67}{12}.⇒sin230°+ cos230°5 cos260°+4 sec230°− tan245°⇒(21)2+(23)25×(21)2+4×(32)2−12⇒41+435×41+4×34−1⇒4445+316−1⇒11215+64−12⇒1267.
Hence, 5 cos260°+4 sec230°− tan245°sin230°+ cos230°=6712\dfrac{5\text{ cos}^2 60° + 4\text{ sec}^2 30° - \text{ tan}^2 45°}{\text{sin}^2 30° + \text{ cos}^2 30°} = \dfrac{67}{12}sin230°+ cos230°5 cos260°+4 sec230°− tan245°=1267.
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cos 45°sec 30° + cosec 30°\dfrac{\text{cos 45°}}{\text{sec 30° + cosec 30°}}sec 30° + cosec 30°cos 45°
sin 30° + tan 45° - cosec 60°sec 30° + cos 60° + cot 45°\dfrac{\text{sin 30° + tan 45° - cosec 60°}}{\text{sec 30° + cos 60° + cot 45°}}sec 30° + cos 60° + cot 45°sin 30° + tan 45° - cosec 60°
Choose the correct option and justify your choice :
2 tan 30°1 + tan230°=\dfrac{\text{2 tan 30°}}{\text{1 + tan}^2 30°} =1 + tan230°2 tan 30°=
sin 60°
cos 60°
tan 60°
sin 30°
1 - tan245°1 + tan245°=\dfrac{\text{1 - tan}^2 45°}{\text{1 + tan}^2 45°} =1 + tan245°1 - tan245°=
tan 90°
1
sin 45°
0