Question

(i) sin o sin(90° - 0) - cos O cos (90° - 0) = 0​

Answer 1

Answer:

given that

let LHS=

sin 0cos (90--0)+sin (90--0)cos0

now using

\begin{gathered}sin(90 - \alpha ) = cos \alpha \\ cos(90 - \alpha ) = sin \alpha \end{gathered}

sin(90−α)=cosα

cos(90−α)=sinα

therefore LHS =sin0sin0+cos0cos0

now using

\begin{gathered} \sin(0 ) = 0 \\ \cos(0) = 1 \end{gathered}

sin(0)=0

cos(0)=1

LHS=0*0+1*1

=1=RHS

Answer 2

Answer:

0

Step-by-step explanation:

sin 0= 0

sin(90-0)= cos0=1

cos0 =1

cos(90-0)=sin0

so, sin 0*sin(90-0) - cos 0 * cos(90-0)

0*1 - 1*0

0-0

0

Hence, Proved