Question

6. Evaluate: cos^2 30° cos^2 45° + 4 sec^2 60° + 1/2 cos^2 90° – 2 tan^2 60°

Answer 1

cos 30 is = √3/2

and cos 45 = 1/√2

sec 60 = 2

tan 60= √3

then we can write

cos^2 30 cos^2 45 + 4 sec^2 60 + 1/2 cos^2 90 - 2 tan^2 60  is now

= (√3/2) ^2 X (1/√2)^2 + 4 X (2)^2 + 1/2 X 0 - 2 X (√3)^2  

= 3/4 X 1/2 + 16 + 0 - 6

= 3/8 + 10

= (3+80)/8

= 83/8

Answer 2

${ \tt { \large \underline {\blue{ solution}}}}$

${ \rm{ {cos}^{2} 30 \degree \: {cos}^{2} 45 \degree +4{sec}^{2} 60 \degree + \dfrac{1}{2} {cos}^{2} 90 \degree - {tan}^{2}90 \degree }}$

${ \rm{ = { \dfrac{ (\sqrt{3} )}{2} }^{2} \times {1}^{2} + 4 \times {2}^{2} + \dfrac{1}{2} \times {0}^{2} - 2 \times { (\sqrt{3} )}^{2} }}$

${ \rm{ = { \dfrac{ 3}{4} } \times {1} + 4 \times 4+ \dfrac{1}{2} \times {0}- 2 \times { {3} } }}$

${ \rm{ = { \dfrac{ 3}{4} } \times {1} + 16+ 0- 2 \times { {3} } }}$

${ \rm{ = { \dfrac{ 3}{4} } + 16 - 6 }}$

${ \rm{ = { \dfrac{ 3}{4} } + 10 }}$

${ \rm{ = \dfrac{3 + 40}{4} }}$

${ \rm{ = \dfrac{43}{4} }}$

${ \rm{ = 10.75}}$

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${ \tt { \large \underline {\blue{ more \: info}}}}$

${ \rm{sin = \dfrac{p}{h} }}$

${ \rm{cos = \dfrac{b}{h} }}$

${ \rm{tan = \dfrac{p}{b} }}$

${ \rm{cosec = \dfrac{h}{p} }}$

${ \rm{sec = \dfrac{h}{b} }}$

${ \rm{sec = \dfrac{b}{p} }}$