2nd sum and 5th sum plzzzzz someone help
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Hey !!!
Q no. 2nd
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Solution :-
from LHS
sin²A cos²B + cos²A sin²B + cos²A cos²B + sin²A sin²B
= sin²A cos²B + cos²A cos²B + cos²A sin²B + sin²A sin²B [ just rearranging the term ]
= cos²B ( sin²A + cos²A ) + sin²B(cos²A + sin²A )
= cos²B × 1 + sin²A × 1 [•°• sin²¢ + cos²¢ = 1 ]
= cos²B + sin²B
= 1 [RHS prooved]
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5th Q no.
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From LHS
tan³¢ /1 + tan²¢ + cot³¢/1+cot²¢
= tan³¢ /sec²¢ + cot³¢/cosec²¢ [•°•1 + cot²¢ = cosec²¢ # and 1 + tan²¢ = sec²¢]
= sin³¢/cos³¢ / 1 /cos²¢ + cos³¢/sin³¢/1 /sin²¢
= sin³¢/cos¢ + cos³¢ /sin¢
= sin⁴¢ + cos⁴¢
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sin¢ × cos¢
=( sin²¢ )² + (cos²¢)²
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sin¢ × cos¢
(a + b)² - 2ab = a² + b² (using this identity )
= (sin²¢ + cos²¢ )²-²sin2¢×cos²¢ /sin¢ ×cos¢
= 1 ² -2sin²¢ cos²¢/sin¢×cos¢
= 1/sin¢Cos¢ - 2sin²¢ cos²¢/sin¢×cos¢
= cosec¢ sec¢ - 2sin¢ cos¢
RHS prooved
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Hope it helps you !!!
@Rajukumar111
Q no. 2nd
_______________________
Solution :-
from LHS
sin²A cos²B + cos²A sin²B + cos²A cos²B + sin²A sin²B
= sin²A cos²B + cos²A cos²B + cos²A sin²B + sin²A sin²B [ just rearranging the term ]
= cos²B ( sin²A + cos²A ) + sin²B(cos²A + sin²A )
= cos²B × 1 + sin²A × 1 [•°• sin²¢ + cos²¢ = 1 ]
= cos²B + sin²B
= 1 [RHS prooved]
_____________
5th Q no.
____________
From LHS
tan³¢ /1 + tan²¢ + cot³¢/1+cot²¢
= tan³¢ /sec²¢ + cot³¢/cosec²¢ [•°•1 + cot²¢ = cosec²¢ # and 1 + tan²¢ = sec²¢]
= sin³¢/cos³¢ / 1 /cos²¢ + cos³¢/sin³¢/1 /sin²¢
= sin³¢/cos¢ + cos³¢ /sin¢
= sin⁴¢ + cos⁴¢
------------------------
sin¢ × cos¢
=( sin²¢ )² + (cos²¢)²
----------------------------
sin¢ × cos¢
(a + b)² - 2ab = a² + b² (using this identity )
= (sin²¢ + cos²¢ )²-²sin2¢×cos²¢ /sin¢ ×cos¢
= 1 ² -2sin²¢ cos²¢/sin¢×cos¢
= 1/sin¢Cos¢ - 2sin²¢ cos²¢/sin¢×cos¢
= cosec¢ sec¢ - 2sin¢ cos¢
RHS prooved
**********************************
Hope it helps you !!!
@Rajukumar111
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