62 - Trigonometry Questions Answers

cosecA+cotA=1/cosecA-cotA

cosecA+cotA=5/2-(1)

cosecA-cotA=2/5-(2)

(1)-(2)=>  

2cotA=5/2-2/5=21/10

=> tanA=20/21

Here, c^2=a^2+b^2 indicates that it is a right angled triangle that means angle C=90degree

and 4s(s-a)(s-b)(s-c)=4(area of triangle)^2   as area of triangle= √s(s-a)(s-b)(s-c)

also area of triangle = 1/2abcosC

2(area of triangle)=ab (cosC=cos90=1)

by squaring on both sides we get

4(area of triangle)^2=a^2b^2 

and thats the required answer.

cosθ.cos(60-θ).cos(60-θ)=1/4cos3θ

=1/4cos3.20.cos60

=1/4.1/2.1/2

=1/16

3(sin²θ+cos²θ)+cos²θ=3(1)+cos²θ

=3+(1+cos2θ)/2

=3+1/2+!/2cos2θ

=7/2+1/2cos2θ

min=7/2-√(1/2)²+0²

      =7/2-1/2=3

max=7/2+√(1/2)²+0²

       =7/2+1/2=4

=-1/cos(B+C)[cosBcosC-sinBsinC]

=-[cos(B+C)]/cos(B+C)

=-1