Answer :

geogebra-export.png


(i) tanABC = cot ACO


AOC = 90° = BOC


CAO = 60° and


ACO = 30°


tan ABC = tan(90 - ACO)


[tan(90 - θ) = cotθ ]


= cot ACO


(ii) sin2BCO + sin2ACO = 1


sin2BCO + sin2ACO = 1


sin2BCO + sin2ACO


sin2BCO + sin2(90 - BCO)


sin2BCO + cos2BCO


= 1


[since, sin2θ + cos2θ = 1]


(iii) cosec2CAB - 1 = tan2ABC


cosec2CAB - 1


cosec(90 - CAB) = secCAB


CAB = ABC


sec2CAB - 1 = tan2 ABC


sec2 ABC – 1 = tan2 ABC


tan2 ABC = tan2 ABC


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

let us find the vWest Bengal Board - Mathematics

let us find the vWest Bengal Board - Mathematics

If <span lang="ENWest Bengal Board - Mathematics

Let us fill up thWest Bengal Board - Mathematics

ABCD is a rectangWest Bengal Board - Mathematics

<span lang="EN-USWest Bengal Board - Mathematics

The value of <imgWest Bengal Board - Mathematics

The value of (sinWest Bengal Board - Mathematics

Let us determine West Bengal Board - Mathematics

Let us determine West Bengal Board - Mathematics