Question

find the value of 4sin 65°/5 cos 25°-13 cos 53°.cosec37° /5(7 sec2 32 ° - 7 cot2 58)​

Answer 1

Answer:

Step-by-step explanation:

Relation between P, B, H:

      2 2 2 2 2 2 H P B , P H B , B H P

, , ,A, B, C Names of angle between Base and Hypo

Ist 2nd 3rd 4th

sin  P / H sin  1/ cosec  tan  sin/ cos sin90    cos

cos  B/ H cosec   1/sin cot  cos/sin cos90    sin

tan  P / B cos  1/sec tan  cot  1 sec90    cosec 

cosec   H / P sec  1/ cos cosec 90    sec

sec  H / B tan  1/ cot tan90    cot 

cot  B/ P cot  1/tan cot90    tan

5th

1. 2 2 sin   cos   1

(i) 2 2 sin   1  cos  (ii) 2 2 cos   1  sin  (iii) 2 sin   1  cos 

(iv) sin   1  cos1  cos (v) 2 cos   1  sin 

(vi) cos   1  sinsec  1

2. 2 2 sec   tan   1

(i) 2 2 sec   1  tan  (ii) 2 2 tan   sec  1 (iii) 2 sec   1  tan 

(iv) 2 tan   sec  1 (v) tan   sec 1sec 1

3. 2 2 cosec   cot   1

(i) 2 2 cosec   1  cot  (ii) 2 2 cot   cosec  1 (iii) 2 cosec    1  cot 

(iv) 2 cot   cosec  1 (v) cot   cosec  1cosec   1