Find the value of
sin²-30⁰ + tan²-45⁰by
sec²23° + cot²-45⁰
Answers
Answer:
The answer is 67/12
Given
\sf \: \dfrac{5 {cos}^{2}60 + 4 {sec}^{2}30 - {tan45}^{2} }{ {sin}^{2} 30 + {cos}^{2}30 }
sin
2
30+cos
2
30
5cos
2
60+4sec
2
30−tan45
2
NoTE
sin²∅ + cos²∅ = 1
tan45 = 1
cos60 = 1/2
sec60 = 2/√3
Now,
(Putting the values)
\begin{gathered} \longrightarrow \: \sf \: \dfrac{5 \times ( \frac{1}{2}) {}^{2} + 4 \times (\frac{2}{ \sqrt{3} }) {}^{2} - {1}^{2} }{1} \\ \\ \longrightarrow \: \sf \: 5 \times \dfrac{1}{4} + 4 \times \dfrac{4}{3} - 1 \\ \\ \longrightarrow \sf \: \dfrac{5}{4} + \dfrac{16}{3} - 1 \\ \\ \longrightarrow \: \sf \: \dfrac{15 + 64 - 12}{12} \\ \\ \longrightarrow \: \sf \: \dfrac{79 - 12}{12} \\ \\ \longrightarrow \sf \: \dfrac{67}{12} \end{gathered}
⟶
1
5×(
2
1
)
2
+4×(
3
2
)
2
−1
2
⟶5×
4
1
+4×
3
4
−1
⟶
4
5
+
3
16
−1
⟶
12
15+64−12
⟶
12
79−12
⟶
12
67
Thus,
\star \: \: \: \boxed{ \boxed{ \sf \: \dfrac{5 {cos}^{2}60 + 4 {sec}^{2}30 - {tan45}^{2} }{ {sin}^{2} 30 + {cos}^{2}30 } = \dfrac{67}{13} }}⋆
sin
2
30+cos
2
30
5cos
2
60+4sec
2
30−tan45
2
=
13
67