Math, asked by momd22294, 19 days ago

1. . Evaluate (111) (1) cos12° - sin78° (1) 1 cosec 31° – sec 599 tan 36° cot 54° (iv) sin 150 sec 750 (vi) tan 26° tan64°​

Answers

Answered by mansichahal02
0

Answer:

cos12°-sin78°

sin(90-12)°-sin78°

cos12°-sin78°

hence proved

may this help u

Answered by amitnrw
0

Given :

(i) cos12° - sin78°

(ii)  cosec 31° – sec 59°

iii) tan 36° - cot 54°

(iv) sin 150° sec 750 °

(vi) tan 26° tan64°​

To Find : Evaluate

Solution:

Sin( 90° - x) = cosx

cos(90° - x) = sinx

Tan (90° - x) = cotx

cot (90° - x) = tanx

tanx = 1/cotx  , cotx = 1/tanx , cotxtanx  = 1

cosec(90° - x) = secx  , sec(90° - x) = cosec x

sinx = 1/cosecx  , cosecx = 1/sinx  ,

cosx = 1/secx  , secx = 1/cosx

(i) cos12° - sin78°

= cos12° - sin(90°-12°)

=  cos12° -  cos12°

= 0

(ii)  cosec 31° – sec 59°

= cosec 31° – sec (90° - 31°)

=  cosec 31° –  cosec 31°

= 0

iii) tan 36°  - cot 54°

=  tan 36° - cot (90° - 36°)

=    tan 36° -   tan 36°

= 0

(iv) sin 150° sec 750 °

= sin ( 180° - 30°) sec( 2 * 360° + 30°)

= sin 30° sec 30°

= sin 30° /cos  30°

= tan 30°

= 1/√3

(vi) tan 26° tan64°​

= tan 26° tan ( 90° - 26°)

= tan 26° cot 26°

= 1

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