Question

8th 9th and 10th sums plzzzz someone help me

Answer 1

hope this give you some idea for question no.8

Answer 2

Hey !!

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
____________
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