Question

Sin e=4/7
find sec²e + cos² e​

Answer 1

Answer:

✥ Question ✥

✽ Sin e=4/7

find sec²e + cos² e

❖ Answer ❖

Given,

Sin e = 4/7

Sin = p/h

So, In the triangle given p = 4 ,h= 7 and

b = ?

✽ Then, by using pythagorean theorem we will find base :

➫ b² = h² - p²

➫ b² = 7² - 4²

➫ b² = 49 - 16

➫ b = √33

✽ Now,

Sec = h / b

= 7/ √33

Cos = b / h

= √33/ 7

A/q,

Sec²e + cos²e

➔ ( 7/√33)² + (√33/7)²

➔ 49/33 + 33/49

➔ 49×33 + 33×49/1617

➔ 1617 + 1617/1617

➔ 3234/1617

➔ 2

✦ Hope it Helps you ! ✦

Answer 2

Answer:

Qᴜᴇsᴛɪᴏɴ

Sin e=4/7

Sin e=4/7find sec²e + cos² e

Sᴏʟᴜᴛɪᴏɴ

ɢɪᴠᴇɴ ,

Sɪɴ e = 4/7

Sin = p/h

In the triangle given p = 4 , h = 7 and b = ?

So, by using Pythagorean theorem we will find the value of base

b² = h² - p²

b² = 7² - 4²

b² = 49 - 16

b = √33

Then,

Sec = h/b = 7/√33

Cos = b/h = √33/7

A/q,

$sec \: {}^{2} e + cos {}^{2} e$

$= > ( \frac{7}{ \sqrt{33} } ) {}^{2} + ( \frac{ \sqrt{33} }{7} ) {}^{2}$

$= > \frac{49}{33} + \frac{33}{49}$

$= > \frac{1617 + 1617}{1617}$

$= > \frac{3234}{1617}$

$= > 2$

Hope it helps you!!!