if sinthitha=3/4,show that√cosec2thitha-cot2thitha/sec2thitha-1=√7/3?
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Step-by-step explanation:
Sin = 3/4 = p/h
P=3k , h=4k (K is constant)
H²=p²+b²
B²=h²-p²
B²=(4k)²-(3k)²
B²=16k²-9k²
B²=7k²
B=/7k
/cosec²-cot²÷sec²-1 =/7÷3
=/(1)² / (tan) ²
=1÷tan
=cot
Cot =b/p
=/7÷3
Lhs = rhs
Hope it works
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Step-by-step explanation: see attachement
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