Question

(tan25°÷cosec65°)^2 +(cot25°÷sec65°)^2 +2 (tan18°×tan45°×tan72°)

Answer 1

hello friends...

here is the answer you wants .

solution:-
shown in attachment

formula used:-
sin²x + cos²x = 1
cosec (90-x) = secx
and sec (90-x) = cosec x
and tan(90-x)= cot x

hence,

(tan25°÷cosec65°)^2 +(cot25°÷sec65°)^2 +2 (tan18°×tan45°×tan72°) = 3 answer

♣♣ hope it helps ♣♣

Answer 2

Answer:

Good night

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