Question

class 10
pls answer
chapter no 8​

Answer 1

Answer:

it will be reported I know but I also have to poste one question so in need points and it is very much important question

Answer 2

Step-by-step explanation:

Given :-

3 Cot A = 4

To find :-

Check whether (1-Tan²A)/(1+Tan²A) = Cos²A- Sin²A or not ?

Solution :-

Given that

3 Cot A = 4

=> Cot A = 4/3

=> 1/Tan A = 4/3

=> Tan A = 3/4

On squaring both sides then

=> Tan² A = 9/16 ------(1)

On adding 1 both sides then

=> 1+Tan²A = 1+(9/16)

=> Sec² A= (16+9)/16

Since, Sec²A - Tan² A = 1

=> Sec² A = 25/16

=> Sec A = √(25/16)

=> Sec A = 5/4

=> 1/Cos A = 5/4

=> Cos A = 4/5

On squaring both sides then

=> Cos² A = 16/25 -----(2)

=> 1-Cos² A = 1-(16/25)

=> Sin²A = (25-16)/25

Since Sin² A + Cos² A = 1

=> Sin² A = 9/25 -----(3)

Now

On taking LHS

(1-Tan²A)/(1+Tan²A)

=> [1-(9/16)] / [ 1+(9/16)]

=> [(16-9)/16]/[(16+9)/16]

=> (7/16)/(25/16)

=> 7/25 ------------(4)

and

On taking RHS

Cos²A- Sin²A

=> (16/25)-(9/25)

=> (16-9)/25

=> 7/25 ----------(5)

From (4)&(5)

LHS = RHS

(1-Tan²A)/(1+Tan²A) = Cos²A- Sin²A

Verified.

Used formulae:-

→ Sec²A - Tan² A = 1

→ Sin² A + Cos² A = 1

→ Cot A = 1/Tan A

→ Sec A = 1/ Cos A