Answer :

Hrishii

Answer:

See Explanation

Step-by-step explanation:

[tex] \csc \theta \: \cos \theta \: \tan \theta = 1 \\ \\ LHS = \csc \theta \: \cos \theta \: \tan \theta \: \\ \\ = \frac{1}{ \sin \theta} \: \times \cancel{ \cos \theta }\: \times \frac{ \sin \theta}{ \cancel{ \cos \theta }} \\ \\ = \frac{1}{ \cancel{ \sin \theta}} \: \times \cancel{ \sin \theta} \\ \\ = 1 \\ \\ = RHS \\ \\ hence \: proved[/tex]

kitkatacpc

Answer:

a) Only the first one is an identity.

Step-by-step explanation:

1). 8 cos O tan O csc O = 8   simplifies to:

cos O tan O csc O = 1

cos O * (sin O / cos O) * (1 /sin O)

=  cos O sin O / cos O sin O

= 1

So it is identity.

2) 13 sec^2 O/ cos^2 O - tan^2 O / cos^2O

= 13 sec^2 O - tan^2 O / cos^2 O

Now  sec^2 O = 1 + tan^2 O, so we have:

(13( 1 + tan^2 O) - tan^2 O) / cos^2 O

= (12 tan^2 O + 13) / cos^2 O

This is not always = 2 so its not an identity.

Step-by-step explanation:

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