L.H.S.

=\(\frac {sin\theta + sin2\theta}{1 + cos\theta + cos2\theta}\)

= \(\frac {sin\theta + 2sin\theta cos\theta}{1 + cos\theta + 2cos^2\theta - 1}\)

= \(\frac {sin\theta (1 + 2cos\theta)}{cos\theta (1 + 2cos\theta)}\)

= \(\frac {sin\theta}{cos\theta}\)

= tan\(\theta\)

Hence, L.H.S. = R.H.S. _proved

Here,

sin3A cos²A

= (3 sinA - 4 sin³A) (1 - sin²A)

= 3 sinA - 4 sin³A - 3 sin³A + 4 sin⁵A

= 4 sin⁵A - 7 sin³A + 3 sinA _Ans

Here,

L.H.S.

= cos 2A

= \(\frac {cos^2A - sin^2A}{1}\)

= \(\frac {cos^2A - sin^2A}{cos^2A + sin^2A}\)

=\(\frac {{\frac {cos^2A}{sin^2A}} - {\frac {sin^2A}{cos^2A}}}{{\frac {cos^2A}{cos^2A}} + {\frac {sin^2A}{cos^2A}}}\)

= \(\frac {1 - tan^2A}{1 + tan^2A}\)

Hence, L.H.S. = R.H.S. _Proved

Here,

tan A = \(\frac 34\)

sin 2A

= \(\frac {2tan A}{1 + tan^2A}\)

= \(\frac {2 × \frac 34}{1 + (\frac {3}{4})^2}\)

= \(\frac {\frac 32}{1 + \frac {9}{16}}\)

= \(\frac 32\)× \(\frac {16}{25}\)

= \(\frac {24}{25}\) _Ans

cos 2A

= \(\frac {1 - tan^2A}{1 + tan^2A}\)

= \(\frac{1 -(\frac {3}{4})^2}{1 + (\frac {3}{4})^2}\)

= \(\frac {1 - \frac 9{16}}{1 +\frac 9{16}}\)

= \(\frac {\frac {16 - 9}{16}}{\frac {16 + 9}{16}}\)

= \(\frac {7}{16}\)× \(\frac {16}{25}\)

= \(\frac 7{25}\) _Ans

L.H.S.

=\(\frac {sin 2A}{1 + cos 2A}\)

= \(\frac {2 sin A cos A}{2 cos^2A}\)

= \(\frac {sin A}{cos A}\)

= tan A

Hence, L.H.S. = R.H.S. _Proved

Here,

cos\(\alpha\) = \(\frac {\sqrt 3}{2}\)

sin\(\alpha\)

= \(\sqrt {1 - cos^2\alpha}\)

= \(\sqrt {1 - (\frac {\sqrt 3}{2})^2}\)

=\(\sqrt {1 - \frac 34}\)

=\(\sqrt {\frac {4 - 3}{4}}\)

=\(\sqrt {\frac 14}\)

= \(\frac 12\)

sin3\(\alpha\)

= 3 sin\(\alpha\) - 4 sin³\(\alpha\)

= 3× \(\frac 12\) - 4× (\(\frac 12\))³

= \(\frac 32\) - \(\frac 48\)

= \(\frac 32\) - \(\frac 12\)

= \(\frac {3 - 1}{2}\)

= \(\frac 22\)

= 1 _Ans

Again,

cos 3\(\alpha\)

= 4 cos²\(\alpha\) - 3 cos\(\alpha\)

= 4 (\(\frac {\sqrt 3}{2}\))³- 3× \(\frac {\sqrt 3}{2}\)

= \(\frac {12\sqrt 3}{8}\) - \(\frac {3\sqrt 3}{2}\)

= \(\frac {3\sqrt 3}{2}\) - \(\frac {3\sqrt 3}{2}\)

= 0 _Ans

L.H.S.

=4 (cos³10° + sin³20°)

=4cos³10° + 4sin³20°

= [cos 3(10°) + 3 cos 10°] + [3 sin20° - sin 3(20°)]

= cos 30° + 3 cos 10° + 3 sin 20° - sin 60°

= \(\frac {\sqrt 3}{2}\) + 3 (cos 10° + sin 20°) - \(\frac {\sqrt 3}{2}\)

=3 (cos 10° + sin 20°)

Hence, L.H.S. = R.H.S. _Proved

Here,

cos\(\theta\) = \(\frac {12}{13}\) and sin\(\alpha\) = \(\frac 45\)

sin\(\theta\)

= \(\sqrt {1 - cos^2\theta}\)

= \(\sqrt {1 - (\frac {12}{13})^2}\)

= \(\sqrt {\frac {169 - 144}{169}}\)

= \(\sqrt {\frac {25}{169}}\)

= \(\frac 5{13}\)

cos\(\alpha\)

= \(\sqrt {1 - sin^2\alpha}\)

= \(\sqrt {1 - (\frac {4}{5})^2}\)

= \(\sqrt {\frac {25 - 16}{25}}\)

= \(\sqrt {\frac 9{25}}\)

= \(\frac 35\)

∴ sin 2\(\theta\) = 2 sin\(\theta\) cos\(\theta\) = 2× \(\frac 5{13}\)× \(\frac {12}{13}\) = \(\frac {120}{169}\) _Ans

∴ cos 2\(\alpha\) = 2 cos²\(\alpha\) - 1 = 2× \(\frac 35\)× \(\frac 35\) - 1 = \(\frac {18 - 25}{25}\) = -\(\frac 7{25}\) Ans

Here,

3\(\theta\) + \(\theta\) = 4\(\theta\)

Putting tan on both,

tan (3\(\theta\) + \(\theta\)) = tan 4\(\theta\)

or, \(\frac {tan 3\theta + tan\theta}{1 - tan 3\theta tan\theta}\) = tan 4\(\theta\)

or, tan 3\(\theta\) + tan\(\theta\) = tan 4\(\theta\) - tan\(\theta\) tan 3\(\theta\) tan 4\(\theta\)

or,tan\(\theta\) tan 3\(\theta\) tan 4\(\theta\) =tan 4\(\theta\) -tan 3\(\theta\) -tan\(\theta\)

Hence, L.H.S. = R.H.S. _proved

L.H.S.

=\(\frac {sin 5\theta}{sin \theta}\) - \(\frac {cos 5\theta}{cos \theta}\)

= \(\frac {sin 5\theta cos \theta - cos 5\theta sin \theta}{sin \theta cos \theta}\)

= \(\frac {sin (5\theta - \theta)}{sin\theta cos\theta}\)

= \(\frac {sin 4\theta}{sin\theta cos\theta}\)

= \(\frac {sin 2(2\theta)}{sin\theta cos\theta}\)

= \(\frac {2 sin 2\theta cos 2\theta}{sin\theta cos\theta}\)

= \(\frac {2 × 2 sin\theta cos\theta cos2\theta}{sin\theta cos\theta}\)

= 4 cos 2\(\theta\)

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

= \(\frac {cos^3A + sin^3A}{cosA + sinA}\)

= \(\frac {(cosA + sinA)(cos^2A + sin^2A - sinA cosA)}{cosA + sinA}\)

= 1 - \(\frac {2 sinA cosA}{2}\)

= 1 - \(\frac 12\)sin 2A

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

=\(\frac 1{tan3A - tanA}\) - \(\frac 1{cot3A - cotA}\)

=\(\frac 1{tan3A - tanA}\) - \(\frac 1{{\frac 1{tan3A}} - {\frac 1{tanA}}}\)

=\(\frac 1{tan3A - tanA}\) - \(\frac {\frac 1{tanA - tan3A}}{tan3A tanA}\)

=\(\frac 1{tan3A - tanA}\) - \(\frac {tan 3A tanA}{tanA - tan3A}\)

= \(\frac 1{tan3A - tanA}\) +\(\frac {tan 3A tanA}{tan3A - tanA}\)

= \(\frac {1 + tan 3A tanA}{tan3A - tanA}\)

= \(\frac {\frac 1{tan3A - tanA}}{1 + tan3A tanA}\)

= \(\frac 1{cot (3A - A)}\)

= \(\frac 1{cot2A}\)

= tan 2A

Hence, L.H.S. = R.H.S. _proved

L.H.S.

=\(\frac {cos\theta}{cos\theta - sin\theta}\) -\(\frac {cos\theta}{cos\theta +sin\theta}\)

= \(\frac {cos\theta (cos\theta + sin\theta) - cos\theta (cos\theta - sin\theta)}{(cos\theta - sin\theta) (cos\theta + sin\theta)}\)

= \(\frac {cos^2\theta + cos\theta sin\theta - cos^2\theta + cos\theta sin\theta}{cos^2\theta - sin^2\theta}\)

= \(\frac {2 sin\theta cos\theta}{cos^2\theta - sin^2\theta}\)

= \(\frac {sin 2\theta}{cos 2\theta}\)

= tan 2\(\theta\)

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

=\(\frac {1 - cos 2A + sin2A}{1 + cos 2A + sin2A}\)

= \(\frac {2sin^2A + 2sinAcosA}{2cos^2A + 2sinAcosA}\) [\(\because\) 1 - cos2A = 2sin²A and 1 + cos2A = 2cos²A]

= \(\frac {2sin A (sinA + cosA)}{2cosA (cosA + sinA)}\)

= tanA

Hence, L.H.S. = R.H.S. _proved

L.H.S.

=\(\frac {1 - tan^2(\frac {π}{4} - \theta)}{1 + tan^2(\frac {π}{4} - \theta)}\)

= cos 2(\(\frac {π}{4} - \theta)\)

= cos(\(\frac {π}{2} - 2\theta)\)

= sin 2\(\theta\)

hence, L.H.S. = R.H.S. _Proved

L.H.S.

=\(\frac {sin 3A}{sinA}\) - \(\frac {cos 3A}{cosA}\)

= \(\frac {3 sinA - 4 sin^3A}{sinA}\) - \(\frac {4 cos^3A - 3 cosA}{cosA}\)

= \(\frac {sinA (3 - 4 sin^2A)}{sinA}\) - \(\frac {cosA (4 cos^2A - 3)}{cosA}\)

= 3 - 4 (1 - cos²A) - 4 cos²A + 3

= 3 - 4 + 4 cos²A - 4 cos²A + 3

= 6 - 4

= 2

Hence, L.H.S = R.H.S. _Proved

L.H.S.

=4 cosec 2A . cot 2A

= 4(\(\frac 1{sin2A})\) . \(\frac {cos 2A}{sin 2A}\)

= \(\frac {4 (cos^2A - sin^2A)}{2 sinA cosA . 2 sinA cosA}\)

= \(\frac {cos^2A - sin^2A}{sin^2A cos^2A}\)

= \(\frac {cos^2A}{sin^2A cos^2A}\) - \(\frac {sin^2A}{sin^2A cos^2A}\)

= \(\frac {1}{sin^2A}\) - \(\frac {1}{cos^2A}\)

= cosec²A - sec²A

Hence, L.H.S. = R.H.S. _proved

Here,

cosA = \(\frac {15}{17}\)

cosA

= \(\sqrt {1 - cos^2A}\)

= \(\sqrt {1 - (\frac {15}{17}})^2\)

= \(\sqrt {1 - \frac {225}{289}}\)

= \(\sqrt {\frac {289 - 225}{289}}\)

= \(\sqrt {\frac {64}{289}}\)

= \(\frac 8{17}\)

∴ tanA = \(\frac {sinA}{cosA}\) = \(\frac {\frac {8}{17}}{\frac {15}{17}}\) = \(\frac 8{15}\)

Now,

tan 3A

= \(\frac {3 tanA - tan^3A}{1 - 3 tan^2A}\)

= \(\frac {3 × \frac 8{15} - (\frac 8{15})^3}{1 - 3 ×(\frac 8{15})^2}\)

= \(\frac {\frac {24}{15} - \frac {512}{3375}}{1 - \frac {192}{225}}\)

= \(\frac {\frac {5400 - 512}{3375}}{\frac {225 - 192}{225}}\)

= \(\frac {\frac {4888}{3375}}{\frac {33}{225}}\)

= \(\frac {4888}{3375}\)× \(\frac {225}{33}\)

= \(\frac {4888}{15 × 33}\)

= \(\frac {4888}{495}\) _Ans

L.H.S.

=\(\frac 1{tan 3A + tanA}\) - \(\frac 1{cot 3A + cotA}\)

= \(\frac 1{tan 3A + tanA}\) - \(\frac 1{{\frac 1{tan 3A}} + {\frac 1{tanA}}}\)

= \(\frac 1{tan 3A + tanA}\) - \(\frac {\frac 1{tanA + tan3A}}{tan3A tanA}\)

= \(\frac 1{tan 3A + tanA}\) - \(\frac {tan 3A tanA}{tan 3A + tanA}\)

= \(\frac {1 - tan 3A tanA}{tan 3A + tanA}\)

= \(\frac {\frac 1{tan 3A + tanA}}{1 - tan3A tanA}\)

= \(\frac 1{tan (3A + A)}\)

= \(\frac 1{tan 4A}\)

= cot 4A

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

=\(\frac {1 + sin 2A}{cos 2A}\)

= \(\frac {sin^2A + cos^2A + 2 sinA cosA}{cos^2A - sin^2A}\)

= \(\frac {(cosA + sinA)^2}{(cosA + sinA) (cosA - sinA)}\)

= \(\frac {cosA + sinA}{cosA - sinA}\)

Hence, L.H.S. = R.H.S. _Proved

Here,

sin 2\(\theta\)

= \(\frac {2 tan\theta}{1 + tan^2\theta}\)

= \(\frac {2 × \frac 43}{1 + (\frac 43)^2}\)

= \(\frac {\frac 83}{1 + \frac {16}{9}}\)

= \(\frac {\frac83}{\frac {9 + 16}{9}}\)

= \(\frac {\frac 83}{\frac {25}9}\)

= \(\frac 83\)× \(\frac 9{25}\)

= \(\frac {24}{25}\) _Ans

Here,

cos A = \(\frac 35\)

sinA

= \(\sqrt {1 - cos^2 A}\)

= \(\sqrt {1 - (\frac 35)^2}\)

= \(\sqrt {1 - \frac 9{25}}\)

= \(\sqrt {\frac {25 - 9}{25}}\)

= \(\sqrt {\frac {16}{25}}\)

= \(\frac 45\)

Now,

sin 2A

= 2 sinA cosA

= 2× \(\frac 45\)× \(\frac 35\)

= \(\frac {24}{25}\) _Ans

L.H.S.

= sin⁴A

= (sin²A)²

=[\(\frac 12\) (1 - cos 2A)]²

= \(\frac 14\) [1 - 2 cos 2A + cos²A]

= \(\frac14\) [1 - 2 cos2A + (\(\frac {1 + cos4A}{2}\))]

= \(\frac 14\) [\(\frac {2 - 4 cos2A + 1 + cos4A}2\)]

= \(\frac 18\)(3 - 4 cos 2A + cos 4A)

Hence, L.H.S. = R.H.S. _Proved

Here,

cosec 2A + cot 4A = cotA - cosec 4A

or, cot 4A + cosec 4A = cotA - cosec 2A

L.H.S.

= cot 4A + cosec 4A

= \(\frac {cos 4A}{sin 4A}\) + \(\frac 1{sin 4A}\)

= \(\frac {cos 4A + 1}{sin 4A}\)

= \(\frac {2 cos^22A - 1 + 1}{2 sin 2A cos 2A}\)

= \(\frac {2 cos^22A}{2 sin 2A cos 2A}\)

= \(\frac {cos 2A}{sin 2A}\)

= \(\frac {2 cos^2A - 1}{2 sinA cosA}\)

= \(\frac {2 cos^2A}{2 sinA cosA}\) - \(\frac 1{2 sinA cosA}\)

= cotA - \(\frac 1{sin 2A}\)

= cotA - cosec 2A

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

=\(\frac {\sqrt 3}{sin 20°}\) - \(\frac 1{cos 20°}\)

= \(\frac {\sqrt 3 cos 20° - sin 20°}{sin 20° cos 20°}\)

= \(\frac {2 (\frac {\sqrt 3}{2} cos 20° - \frac 12 sin 20°)}{sin 20° cos 20°}\)

= \(\frac {2 × 2 (sin 60° cos 20° - cos 60° sin 20°)}{2 sin 20° cos 20°}\)

= \(\frac {4 [sin (60° - 20°)]}{sin 2 × 20°}\)

= \(\frac {4 sin 40°}{sin 40°}\)

= 4

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

=\(\frac {1}{sin 10°}\) - \(\frac {\sqrt 3}{cos 10°}\)

= \(\frac {cos 10° - \sqrt 3 sin 10°}{sin 10° cos 10°}\)

= \(\frac {2 (\frac {1}{2} cos 10° - \frac {\sqrt 3}2 sin 10°)}{sin 10° cos 10°}\)

= \(\frac {2 × 2 (sin 30° cos10° - cos 30° sin 10°)}{2 sin 10° cos 10°}\)

= \(\frac {4 [sin (30° - 10°)]}{sin 2 × 10°}\)

= \(\frac {4 sin 20°}{sin 20°}\)

= 4

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

= cos 5A

= cos (2A + 3A)

= cos 2A cos 3A - sin 2A sin 3A

= (2 cos²A - 1) (4 cos³A - 3 cosA) - 2 sinA cosA (3 sinA - 4 sin³A)

= 8 cos⁵A - 6 cos³A - 4 cos³A + 3 cosA - 6 sin²A cosA + 8 sin⁴A cosA

= 8 cos⁵A - 10 cos³A + 3 cosA - 6 (1 - cos²A) cosA + 8 (1 - cos²A)² cosA

= 8 cos⁵A - 10 cos³A + 3 cosA - 6 cosA + 6 cos³A + 8 cosA (1 - 2 cos²A + cos⁴A)

= 8 cos⁵A - 4 cos³A - 3 cosA + 8 cosA - 16 cos³A + 8 cos⁵A

=16 cos⁵A - 20 cos³A + 5 cosA

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

=tan (\(\frac {\pi}{4} + A\)) + tan(\(\frac {\pi}{4} -A\))

= \(\frac {tan \frac {\pi}{4} + tanA}{1 - tan {\frac {\pi}{4}} tanA}\) +\(\frac {tan \frac {\pi}{4} -tanA}{1 +tan {\frac {\pi}{4}} tanA}\)

= \(\frac {1 + tanA}{1 - 1 × tanA}\) + \(\frac {1 -tanA}{1 +1 × tanA}\)

= \(\frac {(1 + tanA)^2 + (1 - tan A)^2}{(1 - tanA) (1 + tanA)}\)

= \(\frac {1 + 2 tanA + tan^2A + 1 - 2 tanA + tan^2A}{1 - tan^2A}\)

= \(\frac {2 + 2 tan^2A}{1 - tan^A}\)

= \(\frac {2 (1 + tan^2A)}{1 - tan^2A}\)

= \(\frac 2{\frac {1 - tan^2A}{1 + tan^2A}}\)

= \(\frac 2{cos 2A}\)

= 2 sec 2A

Hence, L.H.S. = R.H.S. _Proved

Here,

tan\(\theta\) = \(\frac 56\) and tan\(\beta\) = \(\frac 1{11}\)

tan (\(\theta\) + \(\beta\)) = \(\frac {tan \theta + tan \beta}{1 - tan\theta tan\beta}\)

or,tan (\(\theta\) + \(\beta\)) = \(\frac {\frac 56 + \frac 1{11}}{1 - \frac 56 × \frac 1{11}}\)

or, tan (\(\theta\) + \(\beta\)) = \(\frac {\frac {5 × 11 + 1 × 6}{66}}{\frac {66 - 5}{66}}\)

or,tan (\(\theta\) + \(\beta\)) = \(\frac {55 + 6}{66}\)× \(\frac {66}{61}\)

or,tan (\(\theta\) + \(\beta\)) = \(\frac {61}{66}\)× \(\frac {66}{61}\)

or,tan (\(\theta\) + \(\beta\)) = 1

or,tan (\(\theta\) + \(\beta\)) = tan 45°

or,tan (\(\theta\) + \(\beta\)) = tan\(\frac {\pi^c}{4}\)

∴ (\(\theta\) + \(\beta\)) =\(\frac {\pi^c}{4}\)

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

=cos⁶\(\theta\) + sin⁶\(\theta\)

= (cos²\(\theta\))³ + (sin²\(\theta\))³

= (cos²\(\theta\) + sin²\(\theta\)) (cos⁴\(\theta\) - cos²\(\theta\) sin²\(\theta\) + sin⁴\(\theta\))

= {(cos²\(\theta\))² - cos²\(\theta\) sin²\(\theta\) + (sin²\(\theta\))²}

= {(cos²\(\theta\) + sin²\(\theta\))² - 2 cos²\(\theta\) sin²\(\theta\) - cos²\(\theta\) sin²\(\theta\)}

= 1 - 3 sin²\(\theta\) cos²\(\theta\)

= 1 - 3 . \(\frac 44\) (sin²\(\theta\) cos²\(\theta\))

= 1 - \(\frac 34\) (4 sin²\(\theta\) cos²\(\theta\))

= 1 - \(\frac 34\) (2 sin\(\theta\) cos\(\theta\))²

= 1 - \(\frac 34\) (sin 2\(\theta\))²

= 1 - \(\frac 34\) (sin²2\(\theta\))

= 1 - \(\frac 34\) (1 - cos²2\(\theta\))

= \(\frac {4 - 3 + 3 cos^22\theta}{4}\)

= \(\frac {1 + 3 cos^22\theta}{4}\)

= \(\frac 14\) (1 + 3 cos²2\(\theta\))

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

=\(\frac {2 cos 8\theta + 1}{2 cos\theta + 1}\)

=\(\frac {2 cos 2 × 4\theta + 1}{2 cos\theta + 1}\)

=\(\frac {2 (2cos^2 4\theta - 1) + 1}{2cos\theta + 1}\)

=\(\frac {4 cos^2 4\theta -2 + 1}{2cos\theta + 1}\)

=\(\frac {4 cos^2 4\theta - 1}{2cos\theta + 1}\)

=\(\frac {(2cos 4\theta)^2 - (1)^2}{2cos\theta + 1}\)

=\(\frac {(2 cos 4\theta + 1)(2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(2 cos 2 × 2\theta + 1)(2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {[2 (2cos^2\theta - 1) + 1](2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(4 cos^2\theta - 2 + 1)(2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(4 cos^2\theta - 1)(2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {[(2 cos2\theta)^2 - (1)^2](2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(2 cos 2\theta + 1) (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {[2 × (2 cos^2\theta - 1) + 1] (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(4 cos^2\theta - 2 + 1) (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(4 cos^2\theta - 1) (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {[(2 cos\theta)^2 - (1)^2] (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(2 cos\theta + 1) (2 cos \theta - 1) (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)

=(2 cos\(\theta\) - 1) (2 cos 2\(\theta\) - 1) (2 cos 4\(\theta\) - 1)

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

= cos²A + sin²A cos 2B

= cos²A + sin²A(1 - 2 sin²B)

= cos²A + sin²A - 2 sin²A sin²B

= 1 - 2 sin²A sin²B

= 1 - (1 - cos 2A) sin²B

= 1 - sin²B + sin²B cos 2A

= cos²B + sin²B cos 2A

Hence, L.H.S. = R.H.S. Proved

Here,

cos 4\(\theta\)

= cos 2 (2\(\theta\))

= 1 - 2 sin² (2\(\theta\))

= 1 - 2 (2sin\(\theta\) cos\(\theta\))²

= 1 - 2× 4 sin²\(\theta\) cos²\(\theta\)

= 1 - 8 sin²\(\theta\) (1 - sin²\(\theta\))

= 1 - 8 sin²\(\theta\) + 8 sin⁴\(\theta\) _Ans

Here,

tan²\(\alpha\) = 1 + 2 tan²\(\beta\)

or, \(\frac {1 - cos 2\alpha}{1 + cos 2\alpha}\) = 1 + 2 (\(\frac {1 - cos 2\beta}{1 + cos 2\beta})\)

or,\(\frac {1 - cos 2\alpha}{1 + cos 2\alpha}\) = \(\frac {1 + cos 2\beta + 2 - 2 cos 2\beta}{1 + cos 2\beta}\)

or,\(\frac {1 - cos 2\alpha}{1 + cos 2\alpha}\) = \(\frac {3 - cos 2\beta}{1 + cos 2\beta}\)

or,\(\frac {1 - cos 2\alpha}{1 + cos 2\alpha}\) + 1 = \(\frac {3 - cos 2\beta}{1 + cos 2\beta}\) + 1

or, \(\frac {1 - cos 2\alpha + 1 + cos 2\alpha}{1 + cos 2\alpha}\) = \(\frac {3 - cos 2\beta + 1 + cos 2\beta}{1 + cos 2\beta}\)

or, \(\frac 2{1 + cos 2\alpha}\) = \(\frac 4{1 + cos 2\beta}\)

or, 2 (1 + cos 2\(\beta\)) = 4 (1 + cos 2\(\alpha\))

or, 2 + 2 cos 2\(\beta\) = 4 + 4 cos 2\(\alpha\)

or, 2 cos 2\(\beta\) =4 + 4 cos 2\(\alpha\) - 2

or,2 cos 2\(\beta\) = 2+ 4 cos 2\(\alpha\)

or, cos 2\(\beta\) = \(\frac {2(1 + 2 cos 2\alpha)}{2}\)

or,cos 2\(\beta\) =1 + 2 cos 2\(\alpha\)

∴ L.H.S. = R.H.S. _Proved

L.H.S.

=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + 8 cot 8\(\theta\)

=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + \(\frac 8{tan 8\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + \(\frac 8{tan 2⋅4\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + \(\frac 8{\frac {2 tan 4\theta}{1 - tan^24\theta}}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + \(\frac {8 (1 - tan^24\theta)}{2 tan 4\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac {4 tan^24\theta + 4 (1 - tan^24\theta)}{tan 4\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac {4 tan^24\theta + 4 - 4 tan^24\theta}{tan 4\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac 4{tan 4\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac 4{tan 2⋅2\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac 4{\frac {2 tan 2\theta}{1 - tan^22\theta}}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac {2 (1 - tan^2\theta)}{tan 2\theta}\)

= tan\(\theta\) + \(\frac {2 tan^22\theta + 2 - 2 tan^22\theta}{tan 2\theta}\)

= tan\(\theta\) + \(\frac 2{tan 2\theta}\)

= tan\(\theta\) + \(\frac 2{\frac {2 tan\theta}{1 - tan^2\theta}}\)

= tan\(\theta\) + \(\frac {1 - tan^2\theta}{tan\theta}\)

= \(\frac {tan^2\theta + 1 - tan^2\theta}{tan\theta}\)

= \(\frac 1{tan\theta}\)

= cot\(\theta\)

Hence, L.H.S. = R.H.S. _proved

R.H.S.

= 8 (cos²x - cos²y) (cos²x - sin²y)

= 2 (2 cos²x - 2 cos²y) (2 cos²x - 2 sin²y)

= 2 [1 + cos 2x - (2 + cos 2y)] [1 + cos 2x - (1 - cos 2y)]

= 2 [1 + cos 2x - 2 - cos 2y] [1 + cos 2x - 1 + cos 2y]

= 2 (cos 2x - cos 2y) (cos 2x - cos 2y)

= 2 (cos²2x - cos²2y)

= 2 cos²2x - 2 cos²2y

= 1 + cos 4x - (1 + cos 4y)

= 1 + cos 4x - 1 - cos 4y

= cos 4x - cos 4y

Hence, L.H.S. =R.H.S. _Proved

R.H.S.

= \(\frac 2{\sqrt {2 + \sqrt {2 + 2 cos 4x}}}\)

= \(\frac 2{\sqrt {2 + \sqrt {2 + 2 cos 2 ⋅2x}}}\)

=\(\frac 2{\sqrt {2 + \sqrt {2 (1 + cos 2 ⋅2x)}}}\)

=\(\frac 2{\sqrt {2 + \sqrt {2 ⋅ 2 cos^22x}}}\)

= \(\frac 2{\sqrt {2 + 2 cos 2x}}\)

=\(\frac 2{\sqrt {2 (1 + cos 2x)}}\)

=\(\frac 2{\sqrt {2 ⋅ 2 cos^2x}}\)

= \(\frac 2{2 cosx}\)

= \(\frac 1{cosx}\)

= sec x

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

=4 sin³\(\alpha\)⋅ cos 3\(\alpha\) + 4 cos³\(\alpha\) ⋅sin 3\(\alpha\)

= (3 sin\(\alpha\) - sin3\(\alpha\))⋅ cos 3\(\alpha\) + (3 cos\(\alpha\) + cos 3\(\alpha\))⋅ sin 3\(\alpha\)

= 3 sin\(\alpha\) cos 3\(\alpha\) - sin 3\(\alpha\) cos 3\(\alpha\) + 3 cos\(\alpha\) sin 3\(\alpha\) + sin 3\(\alpha\) cos 3\(\alpha\)

= 3 sin\(\alpha\) cos 3\(\alpha\) + 3 cos\(\alpha\) sin 3\(\alpha\)

= 3 (sin\(\alpha\) cos 3\(\alpha\) + cos\(\alpha\) sin 3\(\alpha\))

= 3 sin(\(\alpha\) + 3\(\alpha\))

= 3 sin 4\(\alpha\)

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

=cos³\(\alpha\) . cos 3\(\alpha\) + sin³\(\alpha\) . sin 3\(\alpha\)

= cos³\(\alpha\) (4 cos³\(\alpha\) - 3 cos\(\alpha\)) + sin³\(\alpha\) (3 sin\(\alpha\) - 4 sin³\(\alpha\))

= 4 cos⁶\(\alpha\) - 3 cos⁴\(\alpha\) + 3 sin⁴\(\alpha\) - 4 sin⁶\(\alpha\)

= 4 (cos⁶\(\alpha\) - sin⁶\(\alpha\)) - 3 (cos⁴\(\alpha\) - sin⁴\(\alpha\))

= 4 {(cos²\(\alpha\))³ - (sin²\(\alpha\))³} - 3{(cos²\(\alpha\))² - (sin²\(\alpha\))²}

= 4 (cos²\(\alpha\) - sin²\(\alpha\)) (cos⁴\(\alpha\) + cos²\(\alpha\) sin²\(\alpha\) + sin⁴\(\alpha\)) - 3 (cos²\(\alpha\) +sin²\(\alpha\))(cos²\(\alpha\) - sin²\(\alpha\))

= 4 cos 2\(\alpha\) {(cos²\(\alpha\))² + (sin²\(\alpha\))² + cos²\(\alpha\) sin²\(\alpha\)} - 3 cos 2\(\alpha\)

= cos 2\(\alpha\) [4 {(cos²\(\alpha\) + sin²\(\alpha\))² - 2 cos²\(\alpha\) sin²\(\alpha\) + cos²\(\alpha\) sin²\(\alpha\)} - 3]

= cos 2\(\alpha\) [4 - 4 cos²\(\alpha\) sin²\(\alpha\) - 3]

= cos 2\(\alpha\) [1 - sin²2\(\alpha\)]

= cos 2\(\alpha\)⋅ cos²2\(\alpha\)

= cos³2\(\alpha\)

∴ L.H.S. = R.H.S. _Proved

R.H.S.

=2 (cos 4\(\theta\) + 1)

= 2 [cos 2. 2\(\theta\)] + 1

= 2 [2 cos² 2\(\theta\) - 1] + 1

= 4 cos² 2\(\theta\) - 2 + 1

= 4 (2 cos²\(\theta\) - 1)² - 1

= 4 (4 cos⁴\(\theta\) - 4 cos²\(\theta\) + 1) - 1

= 16 cos⁴\(\theta\) - 16 cos²\(\theta\) + 4 - 1

= 16 cos⁴\(\theta\) - 16 cos²\9\theta\) + 3

= 16 cos⁴\(\theta\) - 12 cos²\(\theta\) - 4 cos²\(\theta\) + 3

= 4 cos²\(\theta\) (4 cos²\(\theta\) - 3) - 1 (4 cos²\(\theta\) - 3)

= (4 cos²\(\theta\) - 3) (4 cos²\(\theta\) - 1)

=(4 cos²\(\theta\) - 2 - 1) [(2 cos\(\theta\))² - (1)²]

= [2 (2 cos²\(\theta\) -1) - 1] (2 cos\(\theta\) + 1) (2 cos\(\theta\) - 1)

= (2 cos 2\(\theta\) - 1)(2 cos\(\theta\) + 1) (2 cos\(\theta\) - 1)

=(2 cos\(\theta\) + 1) (2 cos\(\theta\) - 1)(2 cos 2\(\theta\) - 1)

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

=2 (cos⁶x + sin⁶x) - 3 (cos⁴x + sin⁴x) + 1

= 2 [(cos²x)³ + (sin²x)³] - 3 [(cos²x)² + (sin²x)²] + 1

= 2 (cos²x + sin²x) [(cos²x)² - cos²x sin²x + (sin²x)²] - 3 [(cos²x + sin²x)² - 2 cos²x sin²x] + 1

= 2× 1 [(cos²x + sin²x)² - 2 cos²x sin²x - cos²x sin²x] - 3 [1 - 2 sin²x cos²x] + 1

= 2 (1 - 3 cos²x sin²x) - 3 + 6 cos²x sin²x + 1

= 2 - 6 cos²x sin²x - 2 +6 cos²x sin²x

= 0

Hence, L.H.S. = R.H.S. _Proved

R.H.S.

= tan (A + B)

= \(\frac {sin (A + B)}{cos (A + B)}\)× \(\frac {sin (A - B)}{sin (A - B)}\)

= \(\frac {(sinA cosB + cosA sinB) (sinA cosB - cosA sinB)}{(cosA cosB - sinA sinB) (sinA cosB - cosA sinB)}\)

= \(\frac {sin^2A cos^2B - cos^2A sin^2B}{sinA cosA cos^2B - sinB cosB cos^2A - sinB cosB sin^2A - cosA sinA sin^2B}\)

= \(\frac {sin^2A (1 - sin^2B) - (1 - sin^2A) sin^2B}{sinA cosA (cos^2B + sin^2B) - sinB cosB (cos^2A + sin^2A)}\)

= \(\frac {sin^2A - sin^2A sin^2B - sin^2B + sin^2A sin^2B}{sinA cosA . 1 - sinB cosB . 1}\)

=\(\frac {sin^2A - sin^2B}{sinA cosA - sinB cosB}\)

Hence, L.H.S. = R.H.S. _Proved

L.H.S.

=cos⁶\(\theta\) - sin⁶\(\theta\)

= (cos²\(\theta\))³- (sin²\(\theta\))³

= (cos²\(\theta\) - sin²\(\theta\)) [(cos²\(\theta\))² + cos²\(\theta\) sin²\(\theta\) + (sin²\(\theta\))²]

= cos 2\(\theta\) [(cos²\(\theta\))² + 2 cos²\(\theta\) sin²\(\theta\) + (sin²\(\theta\))² - cos²\(\theta\) sin²\(\theta\)]

= cos 2\(\theta\) [(cos²\(\theta\) + sin²\(\theta\))² - \(\frac 14\) × 4 cos²\(\theta\) sin²\(\theta\)]

= cos 2\(\theta\) [(1)² - \(\frac 14\) (2 sin\(\theta\) cos\(\theta\))²]

= cos 2\(\theta\) [1 - \(\frac 14\) (sin 2\(\theta\))²]

=cos 2\(\theta\) (1 - \(\frac 14\) sin² 2\(\theta\))

Hence, L.H.S. = R.H.S. _Proved