cot 46°tan 44°−3sec 20°cosec 70°\dfrac{\text{cot 46°}}{\text{tan 44°}} - 3 \dfrac{\text{sec 20°}}{\text{cosec 70°}}tan 44°cot 46°−3cosec 70°sec 20° + 5 is equal to :
-3
4
3
-4
14 Likes
Solving,
⇒cot 46°tan 44°−3sec 20°cosec 70°+5⇒cot (90° - 44°)tan 44°−3sec (90° - 70°)cosec 70°+5⇒tan 44°tan 44°−3cosec 70°cosec 70°+5⇒1−3×1+5⇒1−3+5⇒3.\Rightarrow \dfrac{\text{cot 46°}}{\text{tan 44°}} - 3 \dfrac{\text{sec 20°}}{\text{cosec 70°}} + 5 \\[1em] \Rightarrow \dfrac{\text{cot (90° - 44°)}}{\text{tan 44°}} - 3 \dfrac{\text{sec (90° - 70°)}}{\text{cosec 70°}} + 5 \\[1em] \Rightarrow \dfrac{\text{tan 44°}}{\text{tan 44°}} - 3 \dfrac{\text{cosec 70°}}{\text{cosec 70°}} + 5 \\[1em] \Rightarrow 1 - 3 \times 1 + 5 \\[1em] \Rightarrow 1 - 3 + 5 \\[1em] \Rightarrow 3.⇒tan 44°cot 46°−3cosec 70°sec 20°+5⇒tan 44°cot (90° - 44°)−3cosec 70°sec (90° - 70°)+5⇒tan 44°tan 44°−3cosec 70°cosec 70°+5⇒1−3×1+5⇒1−3+5⇒3.
Hence, Option 3 is the correct option.
Answered By
6 Likes
sin2 A + sin2 (90° - A) is equal to :
-1
1
2
-2
In a triangle ABC, sec C+A2\dfrac{C + A}{2}2C+A is equal to :
cosec B2\dfrac{B}{2}2B
sec B2\dfrac{B}{2}2B
cosec B+A2\dfrac{B + A}{2}2B+A
none of these
sin 67° . cos 23° + cos 67° . sin 23° is equal to :
2 sin 67°
2 cos 23°
cos θ. cos (90° - θ)tan (90° - θ)\dfrac{\text{cos θ. cos (90° - θ)}}{\text{tan (90° - θ)}}tan (90° - θ)cos θ. cos (90° - θ) is equivalent to :
cos2 θ - 1
sin2 θ
sin2 θ - cos2 θ
sin2 θ - 1
