Question

it is maths question from trigonometry​

Answer 1

Answer:

answer =2

Step-by-step explanation:

properties used:

i)sin^2 x+cos^2 x=1

ii)sec^2 x-tan^2 x=1

iii)cosec^2 x=1+cot^2 x

iv)function=(90 - cofunction) .....eg sinx=cos(90-x)..

=> -1/-1+1=2

Answer 2

SOLUTION

TO DETERMINE

$\displaystyle \sf{ \frac{ { \cot}^{2} {66}^{ \circ} - { \sec}^{2} {24}^{ \circ} }{ { \tan}^{2} {43}^{ \circ} - { \csc}^{2} {47}^{ \circ} } + { \sin}^{2} {79}^{ \circ} + { \sin}^{2} {11}^{ \circ} }$

FORMULA TO BE IMPLEMENTED

We are aware of the Trigonometric formula that

$\displaystyle \sf{1. \: \: \: 1 + { \cot}^{2} \theta = { \csc}^{2}\theta}$

$\displaystyle \sf{2. \: \: \: 1 + { \tan}^{2} \theta = { \sec}^{2}\theta}$

$\displaystyle \sf{3. \: \: \: { \sin}^{2} \theta + { \cos}^{2}\theta} = 1$

EVALUATION

$\displaystyle \sf{ \frac{ { \cot}^{2} {66}^{ \circ} - { \sec}^{2} {24}^{ \circ} }{ { \tan}^{2} {43}^{ \circ} - { \csc}^{2} {47}^{ \circ} } + { \sin}^{2} {79}^{ \circ} + { \sin}^{2} {11}^{ \circ} }$

$\displaystyle \sf{ = \frac{ { \cot}^{2} ({90}^{ \circ} - {24}^{ \circ}) - { \sec}^{2} {24}^{ \circ} }{ { \tan}^{2} ({90}^{ \circ} - {47}^{ \circ}) - { \csc}^{2} {47}^{ \circ} } + { \sin}^{2} {79}^{ \circ} + { \sin}^{2}( {90}^{ \circ} - {79}^{ \circ}) }$

$\displaystyle \sf{ = \frac{ { \tan}^{2} {24}^{ \circ}- { \sec}^{2} {24}^{ \circ} }{ { \cot}^{2} {47}^{ \circ} - { \csc}^{2} {47}^{ \circ} } + { \sin}^{2} {79}^{ \circ} + { \cos}^{2}{79}^{ \circ}}$

$\displaystyle \sf{ = \frac{ - 1 }{ - 1 } + 1}$

$\displaystyle \sf{ = 1 + 1}$

$\displaystyle \sf{ =2}$

Note : In the above solution csc means cosec ( For latex )

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