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Question

Without using trigonometric tables, prove that:

(i)tan 48 tan 23 tan 42 tan 67=1

(ii)tan 7 tan 23 tan 60 tan 67 tan 83=3

(iii)cot 12 cot 38 cot 52 cot 60 cot 78=13 [3 MARKS]

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Solution

Solution: 1 Mark each

We have:

(i)tan 48 tan 23 tan 42 tan 67

=(tan 48 tan 42)(tan 23 tan 67)

={tan 48.tan(9048)}{tan 23.tan(9023)}

=(tan 48 cot 48)(tan 23 cot 23)

[tan(90θ)=cot θ]

=1×1=1


(ii)tan 7 tan 23 tan 60 tan 67 tan 83

=tan 7 tan 83.tan 23.tan(9023).tan 60

=(tan 7.cot 7).(tan 23.cot 23).tan 60

[tan(90θ)=cot θ]

=1×1×3=3

[tan θ.cot θ=1 and tan 60=3]


(iii)cot 12 cot 38 cot 52 cot 60 cot 78

=(cot 12.cot 78).(cot 38.cot 52).cot 60

={cot 12.cot(9012)}{cot 38.cot(9038)}.cot 60

=(cot 12.tan 12).(cot 38.tan 38).cot 60

[cot(90θ)=tan θ]

=1×1×13=13 [cot θ.tan θ=1 and cot 60=13]


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