Question

Evaluate without using trigonometric tables
sin2 28° + sin? 62° + tan238° - cot2 52° + sec230°​

Answer 1

Answer:

Step-by-step explanation:

Answer is 4/3.

cos(90-A)=sinA

Therefore sin^2 28 can be written as cos^2 62

(90-28=62)

Therefore the expression becomes cos^2 62+sin^2 62 +tan^2 38 -cot^2 52

(Also cos^2 A+sin^2 A =1)

Therefore equation becomes 1 + tan^2 38 - cot^2 62.

Similarly tan(90-A)=cotA

Therefore tan^2 38= cot^2(90-38)=cot^2 62.

Now the expression is 1 + cot^2 62-cot^2 62 +1/4sec^2 30.

=> 1+1/4 sec^2 30

=>1+1/3[Since sec 30=2/square root(3) sec^2 30=4/3

1/4*sec ^2 30=(1/4)*(4/3)=1/3.]

=>4/3.

Answer 2

Answer:

$\huge\colorbox{yellow}{Answer\:-}$

${ \sin }^{2}28 + { \sin }^{2}62 + { \tan }^{2}38 - { \cot }^{2}52 + \frac{1}{4} { \sec }^{2}30$

$= { \sin }^{2}28 + { \sin }^{2}(90 - 28) + { \tan }^{2}38 - { \cot }^{2}(90 - 38) + \frac{1}{4} { \sec }^{2}30$

$= ( { \sin }^{2}28 + { \cos }^{2}28) + { \tan }^{2}38 - { \tan }^{2}38 + \frac{1}{4} \times {( \frac{2}{ \sqrt{3} } )}^{2}$

$= 1 + 0 + \frac{1}{4} \times \frac{4}{3}$

$= 1 + \frac{1}{3}$

$\pink{ = > \frac{4}{3}}$

$\huge\colorbox{yellow}{Thank\:You}$