if tan thita=1/√5,what is the value of cosec²thita-sec²thita/cosec²thita+sec²thita
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Step-by-step explanation:
tanØ = 1/√5 cotØ = √5
cosec²Ø-sec²Ø/cosec²Ø+sec²Ø
= 1+cot²A-1-tan²A / 1+cot²A+1+tan²A
= cot²A-tan²A / 2+cot²A+tan²A
Putting ,tanØ = 1/√5, cotØ = √5 in cot²A-tan²A / 2+cot²A+tan²A
= (√5)²-(1/√5)² / 2+(√5)²+(1/√5)²
= (5-1/5) / (2+5+1/5)
= (24/5) / (2+26/5)
= (24/5) / (36/5)
= (24/5) × (5/36)
= 24/36
= 2/3
Hope it helps....... Mark it as the Brainliest answer
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