Question

Answer this too....using multiple angle formula.
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Answer 1

Answer:

Step-by-step explanation:

LHS = cos^6 A + sin^6 A

= (cos²A)³ + (sin²A)³

Using identity-:

(a + b)³ = a³ +b³ + 3ab ( a+b)

=> a³ +b³ = ( a+ b)³ -3ab ( a+ b)

So, (cos²A)^3 + ( sin² A)^3

=(cos²A+ sin²A)^3 - 3 cos²A sin²A(cos²A+sin² A)

= 1- 3cos² A sin² A                   ( since sin² A + cos² A = 1 ) ……………….. LHS

Now, RHS = { 1 + 3cos² ( 2A)} / 4

= { 1 + 3 ( cos 2A )² } /4

= {1+ 3 ( cos² A - sin² A )² } /4                 { since cos 2A = cos² A - sin² A) }

= { 1+ 3( cosA + sinA)² ( cosA - sinA)² }/4

= {1+ 3( 1 + 2sinA cosA) ( 1 - 2sinA cosA)} /4

= { 1 + 3 ( 1 - 4 sin² A cos² A) } / 4

= { 1 + 3 - 12 sin² A cos² A} /4

= (4 - 12 sin² A cos² A) /4

=> 1 - 3 sin²A cos ² A ………….. RHS

Hence proved that LHS = RHS

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I am taking the second question as

1/sin10-√3/cos10 = 4

This is because you haven't given the value of "A" in sin2A

1/sin10-√3/cos10

= (cos10 - √3sin10)/cos10sin10

multiplying 1/2 in denominator and numerator

= {(cos10/2) - (√3sin10/2)}/(cos10sin10/2)

= (cos60cos10 - sin60sin10)/(cos10sin10/2)

= cos(60 + 10)/(cos10sin10/2)

= cos70/(cos10sin10/2)

= 2cos70/cos10sin10

multiplying 2 on numerator and denominator

= 4cos70/2sin10cos10

=4cos70/sin20

=4cos70/cos70

= 4                 [proved]

Answer 2

Step-by-step explanation:

Hope the solution is clear. if not, then ask the question again with your specific doubts.