8th 9th and 10th sums plzzzz someone help me
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8th 9th and 10th sums plzzzz someone help me
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Qno. 8
question is wrong .
question should be
sinA /1+ cosA + 1 + cosA/sinA = 2cosecA
FROM LHS
= sin²A + (1 + cosA )²
---------------------------
sinA ( 1 + cosA)
= sin²A + 1² + cos²A + 2cosA
----------------------------------------
sinA ( 1 + cosA )
= sin²A +cos²A +1 + 2cosA
------------------------------------
sinA ( 1 + cosA)
= 1 + 1 + 2cosA
-------------------
sinA ( 1 + cosA)
= 2 ( 1 + cosA )
---------------------
sinA ( 1 + cosA)
= 2/sinA = 2cosecA RHS prooved
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9 th Q no .
_____________
1 + tan²A /1 + cot²A
= sec²A/cosec²A
= 1/cos²A/1 /sin²A = sin²A /cos²A =tan²A
now from RHS
(1 - tanA/1 -cotA)²
=1² + tan²A - 2tanA/1² + cot²A - 2cotA
= sec²A - 2tanA/cosec²A - 2cotA
= sec²A /cosec²A = 2tanA /2cotA
= tan²A = 2tan²A hence lhs = rhs
prooved
___'___________
Q 10
=> cos^6¢ + sin^6¢
= (cos²¢)³ + (sin²¢)³
using identity a³ + b³ = (a+b )a² + b² - ab
= (cos²¢ + sin²¢) ( cos¢)⁴ + (sin¢)⁴ - sin²¢×cos²¢
= 1 ( cos²¢)² + (sin²¢)² - sin²¢*cos²¢
= (cos²¢ + sin²¢ ) ²- 2sin²¢cos²¢ - sin²¢ ×cos²¢
{using this identity , a² + b² = (a + b)² - 2ab }
= 1² - 3sin²¢ cos²¢
= 1 - 3sin²¢ cos²¢ RHS prooved
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Hope it helps you !!!
@Rajukumar11