Question

(cotQ-cosecQ)square =1-CosQ÷1+CosQ
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Answer 1

Answer:

Step-by-step explanation:

TO PROVE:-

$(cotQ - cosecQ)^2 = \frac{1-cosQ}{1+cosQ}$

LHS:-

We know that:-

• cotQ = cosQ/sinQ  

• cosecQ = 1/sinQ

Placing the values, we get:-

$[(\frac{cosQ}{sinQ} )- (\frac{1}{sinQ} )]^2$

$= [\frac{cosQ-1}{sinQ}]^2$

We can also write it as

$\frac{(cosQ-1)^2}{(sinQ)^2}$

$= \frac{(cosQ-1)(cosQ-1)}{(1-cos^2Q)}$

Using a^2-b^2 identity in denominator,_________($a^2-b^2=(a-b)(a+b)$)

$=\frac{(cosQ-1)(cosQ-1)}{(cosQ-1)(cosQ+1)}$

Now (cosQ-1) in numerator and denominator cancels off,

LHS = $\frac{(cosQ-1)}{(cosQ+1)}$

If we take minus common from RHS,

LHS = RHS,

Hence proved.

Answer 2

TO PROVE:

• (cotQ - cosecQ)² = 1 - cosQ/1 + cosQ

SOLUTION:

Taking LHS

Substituting

cotQ = cosQ/sinQ

cosecQ = 1/sinQ

→ ( cosQ/sinQ - 1/sinQ )²

→ [ ( cosQ - 1 )²/sin²Q ]

Simplifying using

( a - b )² = ( a - b )( a - b )

sin²Q = 1 - cos²Q = 1² - cos²Q

Simplifying the formula using

a² - b² = (a + b)(a - b)

sin²Q = ( 1 + cosQ )( 1 - cosQ )

Substituting the values we have

→ [( cosQ - 1 )( cosQ - 1 )]/[( 1 + cosQ )( 1 - cosQ )]

Cancelling the like terms

→ 1 - cosQ/1 + cosQ

Comparing with LHS

1 - cosQ/1 + cosQ = 1 - cosQ/1 + cosQ

LHS = RHS

Hence, proved