Question

If cosecФ-sinФ=m,secФ-cosФ=n.Prove that (m²n)²/³+(mn²)²/³=1[Don't spam]

Answer 1

Answer:

your answer attached in the photo

Answer 2

Note :

Here, @ = Ф ( theta )

Given : -

cosec @ - sin @ = m

sec @ - cos @ = n

Required to prove : -

• (m²n)^⅔ + ( mn² )^⅔ = 1

Identity used : -

sin² @ + cos² @ = 1

Solution : -

cosec @ - sin @ = m

sec @ - cos @ = n

we need to prove that ;

(m²n)^⅔ + ( mn² )^⅔ = 1

Let's consider the given information .

cosec @ - sin @ = m

since,

• cosec @ = 1/ sin@

➜ 1/ sin @ - sin @ = m

➜ 1 - sin@ sin@/ sin@ = m

➜ 1 - sin² @/sin @ = m

From the identity ;

Sin² @ + cos² @ = 1

cos² @ = 1 - sin² @

This implies ;

➜ cos² @/sin @ = m

Hence,

• m = cos² @/ sin @

Similarly,

➜ sec @ - cos @ = n

since ,

• sec @ = 1/cos @

➜ 1/ cos@ - cos @ = n

➜ 1 - cos@ cos@/cos@ = n

➜ 1 - cos² @/ cos @ = n

From the identity ;

Sin² @ + cos² @ = 1

sin² @ = 1 - cos² @

This implies ;

➜ sin² @/cos @ = n

Hence,

• n = sin² @/cos @

Now,

Let's try to prove whether ;

(m²n)^⅔ + ( mn² )^⅔ = 1

Consider the LHS part

(m²n)^⅔ + ( mn² )^⅔

➔ ( [ cos² @/sin @ ]² x sin² @/cos @ )^2/3 + ( cos² @/sin @ x [ sin² @/cos @]² )^2/3

➔ ( cos⁴ @/sin² @ x sin² @/cos @ )^2/3 + ( cos² @/sin @ x sin⁴ @/cos² @ )^2/3

sin² @ , sin² @ get's cancelled because one is in numerator and one is in denominator .

➔ ( cos⁴ @/cos @ )^2/3 + ( cos² @/sin @ x sin⁴ @/cos² @ )^2/3

cos² @ , cos² @ get's cancelled because one is in numerator and one is in denominator .

➔ ( cos⁴ @/cos @ )^2/3 + ( sin⁴ @/sin @ )^2/3

This is simplified as ;

➔ ( cos³ @ )^2/3 + ( sin³ @ )^2/3

➔ cos² @ + sin² @

➔ sin² @ + cos² @

➔ 1 [ from the identity : sin² @ + cos² @ = 1 ]

Consider the RHS part

➔ 1

LHS = RHS

Hence Verified ✓