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grin007 [14]
3 years ago
10

Please tell me the answer with explanation.​

Mathematics
1 answer:
nikitadnepr [17]3 years ago
7 0

Answer:

HELLO DEAR,

GIVEN:-

10sin⁴A + 15cos⁴A = 6

=> 10(sin²A)² + 15cos⁴A = 6

=> 10{(1 - cos²A)²} + 15cos⁴A = 6

=> 10{1 + cos⁴A - 2cos²A} + 15cos⁴A = 6

=> 10 + 10cos⁴A - 20cos²A + 15cos⁴A = 6

=> 25cos⁴A - 20cos²A + 4 = 0

=> 25cos⁴A - 10cos²A - 10cos²A + 4 = 0

=> 5cos²A(5cos²A - 2) - 2(5cos²A - 2) = 0

=> (5cos²A - 2)(5cos²A - 2) = 0

=> cos²A = 2/5

=> cosA = √2/√5 [ secA = √5/√2]

therefore,

sinA = √[1 - 2/5] = √3/√5 [ cosecA = √5/√3]

now,

27cosec^6A + 8sec^6A

=> 27 × {(√5/√3)²}³ + 8 × {(√5/√2)²}³

=> 27 × 125/27 + 8 × 125/8

=> 125 + 125

=> 250.

HENCE, 27cosec^6A + 8sec^6A = 250.

I HOPE IT'S HELP YOU DEAR,

THANKS

Step-by-step explanation:

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y = -2

Step-by-step explanation:

<u>SOLUTION :-</u>

Solve the equation.

  • Add 8 to both sides in order to isolate the equation.

=> y - 8 + 8 = -10 + 8

=> y = -2

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<u>VERIFICATION :-</u>

To verify the solution of y , just put the solution (y = -2) in place of y in equation and check whether L.H.S = R.H.S.

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R.H.S. :-

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Answer:

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A line m is perpendicular to an angle bisector of ∠A. The sides of ∠A intersect this line m at points M and N. Prove that △AMN i
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Answer:


Step-by-step explanation:

<em><u>Given</u></em><u>:</u>       A line m is perpendicular to the angle bisector of ∠A. We call this  

                 intersecting point as D. Hence, in figure ∠ADM=∠ADN =90°.

                 AD is angle bisector of ∠A. Hence, ∠MAD=∠NAD.

<u><em>To Prove</em></u>:   <em><u>ΔAMN is an isosceles triangle. i.e any two sides in ΔAMN are</u></em>

<em>                    </em><em><u>equal. </u></em>

<em><u>Solution</u></em>:  Now, In ΔADM and ΔADN

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                  AD=AD                ...(2) (∵common side)

                  ∠ADM=∠ADN     ...(3) (∵Given)

                  <u><em> Hence, from equation (1),(2),(3) ΔADM ≅ ΔADN</em></u>

                                                         ( ∵ ASA  congruence rule)

                  ⇒<u><em> AM=AN</em></u>

                  Now, In Δ AMN

                 AM=AN (∵ Proved)

                  Hence, ΔAMN is an isosceles  triangle.


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