Give answer for this question and this question is from TRIGONOMETRY PUT PHOTO WITH FULL STEPS
Attachments:
Answers
Answered by
5
(iii) sec 16° cosec 74°- cot 74° tan 16°
⇒cosec(90°-16°) cosec 74°- cot 74° cot(90°-16°) [secθ=cosec(90°- θ),tanθ=cot(90°-θ)]
=cosec 74° cosec 74°- cot 74° cot 74°
=cosec² 74°- cot² 74°
=1(Ans.) [cosec²θ-cot²θ=1]
⇒cosec(90°-16°) cosec 74°- cot 74° cot(90°-16°) [secθ=cosec(90°- θ),tanθ=cot(90°-θ)]
=cosec 74° cosec 74°- cot 74° cot 74°
=cosec² 74°- cot² 74°
=1(Ans.) [cosec²θ-cot²θ=1]
Similar questions