10sin4a + 15cos4a =6
27cosec6a + 8sec6a=
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10sin4 a + 15cos4 a = 6
⇒10sin4 a + 15 ( 1 - sin2 a )2 = 6 ( As we know sin2 a + cos2 a = 1 )
⇒10sin4 a + 15 ( 1 + sin4 a - 2 sin2 a ) = 6
⇒10sin4 a + 15 +15 sin4 a - 30 sin2 a = 6
⇒25sin4 a - 30 sin2 a + 9 = 0
⇒( 5 sin2 a) 2 + ( 3 ) 2 - 2 ( 5 sin2 a) ( 3 ) = 0
⇒( 5 sin 2 a - 3 ) 2 = 0
⇒( 5 sin 2 a - 3 ) = 0
⇒ sin 2 a = 35
⇒ sin a = 35−−√
As we know sin2 a + cos2 a = 1
therefore cos2 a = 1 - sin2 a
cos2 a = 1 - 35
cos2 a = 25
cos a = 25−−√
As we know cosec a = 1sin a = 53−−√ And sec a = 1cos a = 52−−√
For value of 27 cosec 6a + 8 sec 6 a = ?
Substitute values of cosec a and sec a , we get
⇒27 (53−−√)6 + 8 (52−−√)6
⇒27 (53)(53)(53) + 2 (52)(52)(52)
⇒ 125 + 125
⇒ 250
⇒10sin4 a + 15 ( 1 - sin2 a )2 = 6 ( As we know sin2 a + cos2 a = 1 )
⇒10sin4 a + 15 ( 1 + sin4 a - 2 sin2 a ) = 6
⇒10sin4 a + 15 +15 sin4 a - 30 sin2 a = 6
⇒25sin4 a - 30 sin2 a + 9 = 0
⇒( 5 sin2 a) 2 + ( 3 ) 2 - 2 ( 5 sin2 a) ( 3 ) = 0
⇒( 5 sin 2 a - 3 ) 2 = 0
⇒( 5 sin 2 a - 3 ) = 0
⇒ sin 2 a = 35
⇒ sin a = 35−−√
As we know sin2 a + cos2 a = 1
therefore cos2 a = 1 - sin2 a
cos2 a = 1 - 35
cos2 a = 25
cos a = 25−−√
As we know cosec a = 1sin a = 53−−√ And sec a = 1cos a = 52−−√
For value of 27 cosec 6a + 8 sec 6 a = ?
Substitute values of cosec a and sec a , we get
⇒27 (53−−√)6 + 8 (52−−√)6
⇒27 (53)(53)(53) + 2 (52)(52)(52)
⇒ 125 + 125
⇒ 250
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