please give me the answer of no. 13,14
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hence it is given that
root 3 tan theta-1 =0
so
root 3 tan theta = 1
tan theta = 1/root 3
squaring both side
we obtain that
tan ^2 theta = 1/3
sec^2 theta - 1 =1/3
so
sec^2 theta = 1/3 +1
sec^2theta = (1+3)/ 3
sec ^2 theta =4/3
cos ^2 theta =3/4
now according to the question
sin ^2 theta- cos ^2 theta =
1- cos ^2 theta- cos ^2 theta
1-2cos ^2 theta =
1- 2(3/4) [from 1]
1-3/2
=( 2-3)/2
= -1/2
root 3 tan theta-1 =0
so
root 3 tan theta = 1
tan theta = 1/root 3
squaring both side
we obtain that
tan ^2 theta = 1/3
sec^2 theta - 1 =1/3
so
sec^2 theta = 1/3 +1
sec^2theta = (1+3)/ 3
sec ^2 theta =4/3
cos ^2 theta =3/4
now according to the question
sin ^2 theta- cos ^2 theta =
1- cos ^2 theta- cos ^2 theta
1-2cos ^2 theta =
1- 2(3/4) [from 1]
1-3/2
=( 2-3)/2
= -1/2
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