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hodyreva [135]
2 years ago
14

(sin30°+cos60°) (tan60°-sec30°)​

Mathematics
2 answers:
vredina [299]2 years ago
7 0

Answer:

i hope this will help you

olga_2 [115]2 years ago
5 0

Answer:

\frac{ \sqrt{3} }{3}

Step-by-step explanation:

(sin30 +  \cos(60) )( \tan(60)  -  \sec(30)

( \frac{1}{2}  +  \frac{1}{2} )( \sqrt{3} -  \frac{2 \sqrt{3} }{3})

(1)( \sqrt{3} ) over 3

=  \frac{ \sqrt{3} }{3}

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Please someone help me. It's about proving related to transformation of trigonometric ratios. Thank you!!​
kvasek [131]

Answer:

see explanation

Step-by-step explanation:

Using sum to product identities

cos x - cos y = - 2sin(\frac{x+y}{2} )sin(\frac{x-y}{2} ) = 2sin(\frac{x+y}{2} )sin(\frac{y-x}{2} )

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Note that

sin10° = sin(90 - 10)° = cos80°

Thus

\frac{2sin(\frac{10+80}{2})sin(\frac{80-10}{2})  }{2cos(\frac{10+80}{2})cos(\frac{80-10}{2})  } ← cancel 2 from numerator/ denominator

= \frac{sin45}{cos45} × \frac{sin35}{cos35}

= tan45° × tan35° { tan45° = 1 ]

= 1 × tan35°

= tan35° ← as required

5 0
3 years ago
Read 2 more answers
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3 years ago
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Nothing I find in my notes or online doesn't show me how to solve this. I know how to set it up, but brainly ain't letting me su
Anon25 [30]

Answer:

x=8

Step-by-step explanation:

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(divide 8 from 64)

x=8

8 0
2 years ago
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