If tanx-cotx=5 then what is the value of tan²x+cot²x​

27

Step-by-step explanation:

note that tanx cotx = tanx × = 1

Given

tanx - cotx = 5 ( square both sides )

(tanx - cotx)² = 5² = 25 ← expand left side using FOIL

tan²x - 2tanxcotx + cot²x = 25

tan²x - 2 + cot²x = 25 ( add 2 to both sides )

tan²x + cot²x = 27

-4

step-by-step explanation:

do 6-10=

if u need it to be negative.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°