Question

cos³a+sin³a/cosa+sina + cos³a-sin³a/cosa-sina =k​

Answer 1

Question

cos³a+sin³a/cosa+sina + cos³a-sin³a/cosa-sina =k

Answer

= sinA+cosA/sinA-cosA + sinA-cosA/sinA+cosA

= (sinA+cosA)²/(sinA-cosA)(sinA+cosA)

= (sinA-cosA)²/(sinA-cosA)(sinA+cosA)

= sin²A+cos²A+2sinA+ cosA+sin²A+cos²A/ =sin²A-cos²A

= 2/sin²A-cos²A

= 2/sin²A-cos²A = 2/sin²A-(1-sin²A)

=. 2/2sin²A-1

= 2/2(1-cos²A-1)

=2/1-2cos²A

Answer 2

Solution :-

Solving LHS,

→ {(cos³a+sin³a)/(cosa+sina)} + {(cos³a-sin³a)/(cosa-sina)}

using :-

• a³ + b³ = (a + b)(a² + b² - ab)
• a³ - b³ = (a - b)(a² + b² + ab)

→ [{(cosa + sina)(cos²a + sin²a - cos*sina)}/(cosa + sina)] + [{(cosa - sina)(cos²a + sin²a + cos*sina)}/(cosa - sina)]

→ (cos²a + sin²a - cos*sina) + (cos²a + sin²a + cos*sina)

→ cos²a + cos²a + sin²a + sin²a + cos*sina - cos*sina

→ 2cos²a + 2sin²a

→ 2(cos²a + sin²a)

using cos² A + sin² A = 1 ,

→ 2 * 1

→ 2 .

therefore,

→ 2 = k .

Hence, value of k is equal to 2 .

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