In ∆ABC, angle C=90°, then tanA+tanB equals
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SOLUTION.....
In Right Triangle ABC, right angled at C.
Let Angle A be theta then, Angle B be (90-Theta). Here we indicate Theta as a symbol '€'.
tan A + tan B
=tan € + tan ( 90-€)
=tan € + cot €
=tan € + 1/tan €
=tan^€ + 1 ÷ tan €
=sec^€ / tan €
=1/cos € ÷ sin €/ cos €
=1/sin €
=cosec €.
☆☆☆☆HOPE THIS WILL HELP YOU ......
In Right Triangle ABC, right angled at C.
Let Angle A be theta then, Angle B be (90-Theta). Here we indicate Theta as a symbol '€'.
tan A + tan B
=tan € + tan ( 90-€)
=tan € + cot €
=tan € + 1/tan €
=tan^€ + 1 ÷ tan €
=sec^€ / tan €
=1/cos € ÷ sin €/ cos €
=1/sin €
=cosec €.
☆☆☆☆HOPE THIS WILL HELP YOU ......
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romjenc:
tan(90-B) is cotB.
Answered by
9
Answer:
Hopes it helps uuuu....
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