Question

trigonometry identity (without using trigonometry tables )

Answer 1

{tan(90°-65°) / cosec 65°}²+ {cot(90°-65°)/sec 65°} + 2 tan 18° tan 72° tan 45°
{cot 65° / cosec 65°}²+ {tan 65° / sec 65°}²
+ 2 tan(90°-72°) tan 72° (1)
(cot²65°/cosec²65° )+ (tan²65°/sec²65°)
+ 2 (1/tan 72°) tan 72°
(cos²65°/sin⁴65°)+ (sin²65°/cos⁴65°)+2
= {cos²65°(cos⁴65°) + sin²65°(sin⁴65°)}/sin²65°•cos²65° +2
= {cos⁴65°/sin²65°}+{sin⁴65°/cos²65°}+2
= cot²65°+tan²65°+2

Answer 2

Answer:

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}$

TrigonometryTable