4(sec^2 59°-cot^2 31°)÷3-2÷3sin90°+3tan^2 56°×tan^2 34°=x÷3.find the value of `x`?

Question

Deepak Kumar Shringi · Answer

sec 2 59=1+tan 2 59 and tan59=cot31 and tan 56 tan34=1 4/3-2/3+3=11/3 so x=11

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4(sec^2 59°-cot^2 31°)÷3-2÷3sin90°+3tan^2 56°×tan^2 34°=x÷3.find the value of `x`?

4(sec^2 59°-cot^2 31°)÷3-2÷3sin90°+3tan^2 56°×tan^2 34°=x÷3.find the value of `x`?

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4(sec^2 59°-cot^2 31°)÷3-2÷3sin90°+3tan^2 56°×tan^2 34°=x÷3.find the value of `x`?

4(sec^2 59°-cot^2 31°)÷3-2÷3sin90°+3tan^2 56°×tan^2 34°=x÷3.find the value of `x`?

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