((cosec A - cot A)2 +1)/
(sec A (cosec A - cot A))
= 2 cot A
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Qᴜᴇsᴛɪᴏɴ :-
Prove :- {(cosecA - cotA)² +1}/(secA(cosecA - cotA) = 2cotA
Sᴏʟᴜᴛɪᴏɴ :-
Taking LHS,
→ {(cosecA - cotA)² +1}/(secA(cosecA - cotA)
Putting 1 = cosec²A - cot²A in Numerator,
→ {(cosecA - cotA)² +(cosec²A - cot²A)}/(secA(cosecA - cotA)
using a² - b² = (a + b)(a - b) Now,
→ {(cosecA - cotA)² +(cosecA - cotA)(cosecA + cotA)}/(secA(cosecA - cotA)
Taking (cosecA - cotA) common from Numerator,
→ (cosecA - cotA){(cosecA - cotA) + (cosecA + cotA)} / secA(cosecA - cotA)
→ {(cosecA - cotA) + (cosecA + cotA)} / secA
→ 2cosecA / secA
Putting cosecA = (1/sinA) & secA = (1/cosA) Now,
→ 2*(1/sinA) / (1/cosA)
→ (2/sinA) * (cosA/1)
→ (2cosA/sinA)
→ 2cotA = RHS (Proved).
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QUESTION:
ANSWER:
We use the trigonometry identity here;
other identity we used here :
NOW COME TO YOUR QUESTION :
TAKING LHS;
I LET A AS ALPHA
using the identity
TAKING common here
LHS = RHS
HENCE PROVED
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