Question

2) Given that Sin = 2 2 and
then the value of +B is
Cos B = 1/2
Given that since it and cos Bt. then the value
of a tß is​

Answer 1

ANSWER

ANSWERsinθ+cscθ=2

ANSWERsinθ+cscθ=2⇒sinθ+

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1)

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2 θ+csc

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2 θ+csc 2

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2 θ+csc 2 θ=sin

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2 θ+csc 2 θ=sin 2

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2 θ+csc 2 θ=sin 2 θ+

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2 θ+csc 2 θ=sin 2 θ+ sin

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2 θ+csc 2 θ=sin 2 θ+ sin 2

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2 θ+csc 2 θ=sin 2 θ+ sin 2 θ

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2 θ+csc 2 θ=sin 2 θ+ sin 2 θ1

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2 θ+csc 2 θ=sin 2 θ+ sin 2 θ1

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2 θ+csc 2 θ=sin 2 θ+ sin 2 θ1

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2 θ+csc 2 θ=sin 2 θ+ sin 2 θ1 =1+1

ANSWERsinθ+cscθ=2⇒sinθ+ sinθ1 =2⇒sin 2 θ−2sinθ+1=0⇒(sinθ−1) 2 =0⇒sinθ=1⇒sin 2 θ+csc 2 θ=sin 2 θ+ sin 2 θ1 =1+1=2

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