cos8A.cos5A-cos12A.cos9A / sin8A.cos5A+cos12A.sin9A=
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(cos8A.cos5A-cos12A.cos9A)/(sin8A.cos5A+cos12A.sin9A)
=> [1/2 {cos13A+cos3A}-1/2 {cos21A+cos3A}]/[1/2 {sin13A+sin3A}+1/2 {sin21A-sin3A}]
=>( cos13A-cos21A)/(sin13A+sin21A)
=> 2sin17A.sin4A/2sin17A.cos4A
=> tan4A
=> [1/2 {cos13A+cos3A}-1/2 {cos21A+cos3A}]/[1/2 {sin13A+sin3A}+1/2 {sin21A-sin3A}]
=>( cos13A-cos21A)/(sin13A+sin21A)
=> 2sin17A.sin4A/2sin17A.cos4A
=> tan4A
abhi178:
please mark as brainliest
how it will come 4A in 3rd step
can u explain me
reply me
I have use formula cosA-cosB=2sin (A+B)/2.sin (B-A)/2
and sinA+sinB=2sin (A+B)/2.cos (A-B)/2
ok
tnq
which class r u
may i know where r u from
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4
Answer:
thanks for asking this question same question even I was searching for
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