Question

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Answer 1

Step-by-step explanation:

15) To prove (sin x + cos x)² = 1 + 2 sin x cos x

LHS: (sin x + cos x)² =>

=> sin²x + cos²x + 2 sin x cos x [ (a + b)² = a² + b² + 2ab ]

=> 1 + 2 sin x cos x [ sin²x + cos²x = 1 ]

LHS = RHS hence proved

17) tan²x / sec x + 1 = sec x - 1

LHS: tan²x / sec x + 1 =>

=> (sec²x - 1)/(sec x + 1) [ tan²x + 1 = sec²x ]

=> (sec x + 1)(sec x - 1) / (sec x + 1) [ a²-b² =(a+b)(a-b) ]

=> sec x - 1

LHS = RHS hence proved

19) cot x + tan x = cosec x • sec x

LHS: cot x + tan x =>

=> (cos x/sin x) + (sin x/cos x)

=> (cos²x + sin²x)/(sin x cos x) [ cos²x + sin²x = 1]

=> 1 /sin x•cos x

=> 1/sin x • 1/cos x

=> cosec x • sec x

LHS = RHS hence proved

21) cos⁴x - sin⁴x = 1 - 2sin²x

LHS: cos⁴x - sin⁴x =>

=> (cos²x)² - (sin²x)²

=> (cos²x + sin²x)(cos²x - sin²x)

=> 1•(cos²x - sin²x)

=> cos²x - sin²x

=> 1 - sin²x - sin²x

[ cos²x + sin²x = 1 => cos²x = 1 - sin²x ]

=> 1 - 2 sin²x

LHS = RHS hence proved

Answer 2

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