Question

tan²A + cot²A = sec²A.cosec²A + k, find k.​

Answer 1

Solution:- from LHS :- tan²A + cot²A

∵sin²A/cos²A = tan²A and cos²A/sin²A = cot²A

⟹ ∴ sin²A /cos²A + cos²A /sin²A

⟹ sin²A * sin²A + cos²A * cos²A / sin²A * cos²A

⟹ sin⁴A + cos⁴A / sin²A *cos²A

we know that , a²+b² (a+b)² - 2ab

similarly let sin²A = a and sin²B = b

⟹ [(sin²A)+ (cos²B)]² - 2sin²A *cos²A

⟹( sin²A + cos²A)² - 2sin²A cos²A / sin²A cos²A

⟹ ∵ sin²A + cos²A = 1

⟹ 1 - 2sin²A * cos²A / sin²A * cos²A

⟹ 1/sin²A * cos²A - 2sin²A *cos²A /sin²A * cos²A

⟹ cosec²A *sec²A - 2 ______LHS

∵ 1//sin²A = cosec²A and 1/cos²A = sec²A

Compare to given R.H.S

sec²A* cosec²A + k = cosec²A * sec²A -2

We get ,

• k = -2 Answer

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Answer 2

Step-by-step explanation:

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