Without using trigonometric tables, evaluate the following:
cosec 17°30′sec 72° 30′\dfrac{\text{cosec 17°30}'}{\text{sec 72° 30}'}sec 72° 30′cosec 17°30′.
34 Likes
Solving,
⇒cosec 17° 30′sec 72° 30′⇒cosec (90° - 72° 30′)sec 72° 30′As, cosec (90 - θ) = sec θ⇒sec 72° 30′sec 72° 30′⇒1.\Rightarrow \dfrac{\text{cosec 17° 30}'}{\text{sec 72° 30}'} \\[1em] \Rightarrow \dfrac{\text{cosec (90° - 72° 30}')}{\text{sec 72° 30}'} \\[1em] \text{As, cosec (90 - θ) = sec θ} \\[1em] \Rightarrow \dfrac{\text{sec 72° 30}'}{\text{sec 72° 30}'} \\[1em] \Rightarrow 1.⇒sec 72° 30′cosec 17° 30′⇒sec 72° 30′cosec (90° - 72° 30′)As, cosec (90 - θ) = sec θ⇒sec 72° 30′sec 72° 30′⇒1.
Hence, cosec 17° 30′sec 72° 30′=1\dfrac{\text{cosec 17° 30}'}{\text{sec 72° 30}'} = 1sec 72° 30′cosec 17° 30′=1.
Answered By
17 Likes
cos 18°sin 72°\dfrac{\text{cos 18°}}{\text{sin 72°}}sin 72°cos 18°
tan 41°cot 49°\dfrac{\text{tan 41°}}{\text{cot 49°}}cot 49°tan 41°
cot 40°tan 50°−12(cos 35°sin 55°)\dfrac{\text{cot 40°}}{\text{tan 50°}} - \dfrac{1}{2}\Big(\dfrac{\text{cos 35°}}{\text{sin 55°}}\Big)tan 50°cot 40°−21(sin 55°cos 35°)
(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