Question

√2(cos25-sin25)/sin20 1-cos^2 13/cos 26 1-2sin^2 46/8cos92

Answer 1

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Answer 2

To simplify the given expression, let's first manipulate each part individually using trigonometric identities:

√2(cos25-sin25)/sin20

= √2(cos(45-20) - sin(45-20)) / sin20 [using the identity: sin(a-b) = sin(a)cos(b) - cos(a)sin(b), and a=45, b=20]

= √2(cos45cos20 - sin45sin20 - cos45sin20 + sin45cos20) / sin20 [using the identity: cos(a-b) = cos(a)cos(b) + sin(a)sin(b), and a=45, b=20]

= √2[(cos45cos20 - cos45sin20) - (sin45sin20 - sin45cos20)] / sin20

= √2[(cos(45-20)) - (sin(45-20))] / sin20 [using the identity: cos(a-b) = cos(a)cos(b) + sin(a)sin(b), and a=45, b=20]

= √2(cos25+sin25)/sin20

Next,

1-cos^2 13/cos 26

= sin^2 13 / cos 26

Finally,

1-2sin^2 46/8cos92

= 1 - (sin46 / 2cos92)^2 [dividing both numerator and denominator by 4]

= (2cos92 / 2cos92) - (sin46 / 2cos92)^2

= (2cos92 - sin^2 46) / 2cos92

Substituting these simplifications back into the original expression, we get:

√2(cos25+sin25)/sin20 * sin^2 13 / cos 26 * (2cos92 - sin^2 46) / 2cos92

Now, we can simplify further by canceling out common factors and using trigonometric identities:

√2(cos25+sin25) * sin^2 13 * (2cos92 - sin^2 46) / (sin20 * cos26 * 2cos92)

= √2(cos25+sin25) * sin^2 13 * (2cos^2 46 + cos^2 46 - sin^2 46) / (sin20 * cos26 * 2cos92)

= √2(cos25+sin25) * sin^2 13 * (cos^2 46 + cos^2 46) / (sin20 * cos26 * 2cos92)

= √2(cos25+sin25) * sin^2 13 * (2cos^2 46) / (sin20 * cos26 * 2cos92)

= √2(cos25+sin25) * sin^2 13 * (cos46 / sin20) * (1 / cos26)

= √2(cos25+sin25) * (sin13 / cos46) * (sin13 / sin20) * (1 / cos26)

= √2(cos25+sin25) * tan13 * (1 / sin20) * (1 / cos26)

= (√2 / sin20) * (cos25tan13 + sin25tan13) * (1 / cos26)

= (√2 / sin20) * (sin25tan13 / cos25tan13 + 1) * (1 / cos26)

= (√2 / sin20) * ((sin25/cos25) / (tan13